https://tuhsphysics.ttsd.k12.or.us/wiki/api.php?action=feedcontributions&user=WikiSysop&feedformat=atomTuHSPhysicsWiki - User contributions [en]2024-03-28T12:04:25ZUser contributionsMediaWiki 1.28.2https://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Main_Page&diff=646Main Page2008-01-10T22:15:11Z<p>WikiSysop: </p>
<hr />
<div><big>Welcome to the Tualatin High School Physics Wiki</big><br />
<br />
<br />
<br />
IB Physics - Douglas Giancoli's Physics - 5th edition<br />
:[[Giancoli Physics (5th ed) Solutions | Solutions to Problems]]<br />
<br />
<br />
[http://tuhsphysics.ttsd.k12.or.us/Tutorial/PreQuiz/index.htm Review for the Skill Set Final]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Main_Page&diff=645Main Page2008-01-10T22:13:52Z<p>WikiSysop: </p>
<hr />
<div><big>Welcome to the Tualatin High School Physics Wiki</big><br />
<br />
<br />
<br />
IB Physics - Douglas Giancoli's Physics - 5th edition<br />
:[[Giancoli Physics (5th ed) Solutions | Solutions to Problems]]<br />
<br />
<br />
[http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/index.htm Review for the Skill Set Final]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Main_Page&diff=644Main Page2008-01-10T22:13:20Z<p>WikiSysop: </p>
<hr />
<div><big>Welcome to the Tualatin High School Physics Wiki</big><br />
<br />
<br />
<br />
IB Physics - Douglas Giancoli's Physics - 5th edition<br />
:[[Giancoli Physics (5th ed) Solutions | Solutions to Problems]]<br />
<br />
<br />
[http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/index.htm | Review for the Skill Set Final]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Blank_Problem_Template&diff=301Blank Problem Template2007-09-11T22:24:24Z<p>WikiSysop: </p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
===1.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
<blockquote><br />
Solution goes here<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_16&diff=300Giancoli Physics (5th ed) Chapter 162007-09-11T22:18:55Z<p>WikiSysop: Reverted edits by Cmurray (Talk); changed back to last version by WikiSysop</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
===1.How many electrons make up a charge of -30.0 <math>\mu</math>C?===<br />
:Since electrons have charges of –1.602 x 10<sup>-19</sup> C all we need to do is divide the given charge by the electron charge to give us the number of electrons. Also, be careful with the whole micro-coulombs thing.<br />
:-30.0 <math>\mu</math>C = -30.0 x 10<sup>-6</sup> C<br />
:(-30.0 x 10<sup>-6</sup> C) /(-1.602 x 10<sup>-19</sup> C) = 1.88 E14 electrons<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(2.7 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. Two charged balls are 20.0 cm apart. They are moved, and the force on each of them is found to have tripled. How far apart are they now?===<br />
:Solving this problem is nice, because in this case we can use the formula <math>F = \frac{kq_1q_2}{r^2}</math> We can set two equations equal to one another. <br />
:<br />
::<math>\frac{3kq_1q_2}{r_1^2} = \frac{kq_1q_2}{r_2^2}</math><br />
:Where r<sub>1</sub> is 0.20 m, and we are trying to solve for r<sub>2</sub><br />
:If we divide both sides by kq<sub>1</sub>q<sub>2</sub> and plug in for r<sub>1</sub> we get:<br />
::<math>\frac{3}{0.20^2} = \frac{1}{r_2^2}</math><br />
:Finally, <br />
:r<sub>2</sub> = .012 m or 11.5 cm<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4.===<br />
:(0.500 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5.What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10<SUP>-12</SUP> m?===<br />
<BLOCKQUOTE> <br />
<P>This is a straight up F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
problem from the hood. <br />
It’s not too bad, the only trick here is the nucleus charge and to solve that <br />
problem we simply multiply the first charge by +26 e. Plug in the values you <br />
must, young Jedi.</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>F = (8.99E9) * (26 * 1.602E-19 * 1.602E-19) / <br />
(1.5E-12)<SUP>2</SUP></P> <br />
<P><STRONG>F = .0027 N or 2.7E-3 N</STRONG></P> <br />
</BLOCKQUOTE><br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6.===<br />
:(9.2 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===7.What is the magnitude of the force a +15 - mC charge exerts on a +3.0-mili coulombs charge 40 cm away? (1 <math>\mu</math>C=10<SUP>-6</SUP> C, 1 mC=10<SUP>-3</SUP>C.)?===<br />
<br />
<BLOCKQUOTE> <br />
<P>Questions like these are ones you cross your fingers for on <br />
the test. Watch out for the whole mili- and mico- coulomb thing and remember <br />
that two positive charges repel. Other than that you can pound this question <br />
out with speed.</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>F = (8.99E9) * (15E-6) * (3E-3) / (.4)<SUP>2</SUP></P> <br />
<P><STRONG>F = 2.5E3 N</STRONG><BR><BR> <br />
<br />
</P><br />
</BLOCKQUOTE><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===8.===<br />
:(3.8x10<sup>14</sup> electrons, 3.4x10<sup>-16</sup> kg)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===9. Imagine space invaders could deposit extra electrons in equal amounts on the earth and now your car, which has mass of 1050 kg. note that the rubber tires would provide some insulation. How much charge Q would need to be placed on your car (same amount on earth) in order to levitate it (overcome gravity) (hint : assume that earth charge is spread uniformly so it acts as if it were located at the earth's center, and then the separation distance is the radius of the earth.)?===<br />
<BLOCKQUOTE> <br />
<P>Ok, seriously man. The author had a little too much time on <br />
his hands to write this problem. Buts it’s assigned so lets get to <br />
it.<BR>Alright, so the first step is to figure out exactly how much force is <br />
required to move your car. Using gravity as the acceleration to counteract, <br />
remember these guys are levitating your car, we can use F = ma.</P> <br />
<P>F = (1050 kg) (9.8 ms-2) = 10,290 N</P> <br />
<p>Now, your car is on the surface of the earth, so the center to center <br />
distance is the radius of the earth essentially, or 6.38 x 10<sup>6</sup> <br />
m.; Setting the coulombic repulsion equal to the force of gravity on the <br />
car we have:<br> <br />
<br> <br />
10,290 N = kq<sub>1</sub>q<sub>2</sub>/r<sup>2<br> <br />
<br> <br />
</sup>Now, since the two charges must be the same (read the problem), this <br />
reduces to:<br> <br />
<br> <br />
10,290 N = kq<sup>2</sup>/r<sup>2<br> <br />
<br> <br />
</sup>10,290 N = (8.99 x 10<sup>9</sup> Nm<sup>2</sup>/C<sup>2</sup>)<b>q<sup>2</sup></b>/(6.38x10<sup>6</sup>m)<sup>2</sup></p> <br />
<P><strong>q = 6825.7 C = <u>6.8e3 C</u></strong><u><BR></u><BR> <br />
<br />
</P></BLOCKQUOTE><br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10.===<br />
:(-5.4x10<sup>7</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. Particles of charge +70, +48, and +80 mC are placed in a line (fig. 16-37) the center one is 0.35m from each of the others. Calculate the net force on each charge due to the other two.===<br />
<div style="text-align: center;"><br />
[[Image:gp5_16_16_37.JPG]]<br />
</div><br><br />
<br />
:<P>First of all, you should have already sketched a diagram of <br />
:the figure because it makes things much, much easier. This problem really <br />
:isn’t that difficult, because we don’t have to deal with angles and instead <br />
:it is <br />
:just plain tedious. Solving this requires you to calculate both forces exerted <br />
:on each particle, determining whether that force is either attractive (like <br />
:me) or repulsive (like my stupid jokes). I will show how to use all the <br />
:necessary calculations for the +70 charge and after that, only the calculated <br />
:values.</P> <br />
:<P>(a) The force on the; +70E-6:;<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP></P> <br />
:<P>+70 and +48:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= <br />
:(8.99E9)(70E-6)(48E-6) / (.35)<SUP>2</SUP> = 246.6 N (left - the 48 repels the 70)</P> <br />
:<P>+70 and –80:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= <br />
:(8.99E9)(70E-6)(-80E-6) / (.7)<SUP>2</SUP> = 102.7 N (right - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:102.7 (right) - 246.6 (left) = <STRONG>-144 N <br />
:(left)</STRONG></P> <br />
:<P>(b) The force on the middle one (The 48E-6 C)<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP></P> <br />
:<P>+48 and +70 :<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(48E-6)(70E-6) / (.35)<SUP>2</SUP> = 246.6 N (right - the 70 repels the 48)</P> <br />
:<P>+48 and –80:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(48E-6)(80E-6) / (.35)<SUP>2</SUP> = 281.8 N (right - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:246.6 N (right) + 281.8 N (right) = <STRONG>528 N (right)</STRONG><BR></P> <br />
:<P>(c) The force on the right one (The -80E-6 C)<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:</P><P>-80 and +48 :<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(80E-6)(48E-6) / (.35)<SUP>2</SUP> = 281.8 N (left - attracted</P> <br />
:<P>-80 and +70:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(80E-6)(70E-6) / (.70)<SUP>2</SUP> = 102.7 N (left - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:281.8 N (left) + 102.7 N (left) = <strong>385 N (left)</strong></P> <br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===12.Three positive particles of charge 11.0 <math>\mu</math>C are located at the corners of an equilateral triangle of side 15.0 cm (Fig. 16-38).; Calculate the magnitude and direction of the net force on each particle.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5_16_16_38.JPG]]<br />
</div><br />
:<P>This problem seems hard because it seems like we’re going to <br />
:end up finding three different answers and plus, they’re not in a straight <br />
:line. However, all we need to do is calculate the magnitude and net force on <br />
:one particle because we are dealing with an equilateral triangle. To solve the <br />
:problem, we need to find the force one particle exerts on the <br />
:other.</P> <br />
:<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = <br />
:(8.99E9)(11E-6)2 / (.15)<SUP>2</SUP> = 48.35 N</P> <br />
:<P>Now, because this is an equilateral triangle, we can assume <br />
:the angle that each particle is exerting on the other is 60<sup>o</sup>. Since the x-axis <br />
:displacement of each particle is the opposite of the other, they cancel one <br />
:another out and all we need to know is the y-axis displacement. To find it we <br />
:use good old Mr. Sine.</P> <br />
:<P>(48.35)(sin 60) = 41.87 * 2 (remember there are two different <br />
:particles acting on each) =</P> <br />
:<P><STRONG>83.7 N (away from the center)</STRONG><BR><BR><br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A charge of 6.00 mC is placed at each corner of a square 1.00 m on a side.; Determine the magnitude and direction of the force on each charge.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P13.jpg]]<br />
</div><br />
<blockquote><br />
Make sure you make a diagram and draw arrows as to where the <br />
force is going. Use F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
and calculate the magnitude and impact <b>each</b> (meaning all 3) charge has <br />
on the other.</P> <br />
<P>First find the force of the two particles that are not on a <br />
diagonal. Since each particle produces and equal force, the charge will be <br />
propelled diagonally, at an angle of 45 degrees.;</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = (8.99E9)(6E-3)<SUP>2</SUP>; <br />
/ (1)<SUP>2</SUP> = 323640 N</P> <br />
<P>323640<SUP>2</SUP> +323640<SUP>2</SUP>; = c<SUP>2</SUP> ; <br />
</P> <br />
<P>c = 457696 N <br />
</P> <br />
<P>Now, we solve for the third charge on the diagonal. <br />
</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = <br />
(8.99E9)(6E-3)<SUP>2</SUP>; / 2 <br />
</P> <br />
<P>F = 161820 N <br />
</P><P>Now just add the two and the answer is yours. <br />
</P><P>F = 161820 + 457696 = 619516 N = <b>6.2 E5 N</b> <br />
</P> <br />
<P><br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===14.===<br />
2.96x10<sup>5</sup> N Toward the center of the square<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15.Compare the electric force holding the electron in orbit around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton.; What is the ratio of the these two forces?===<br />
<br />
<blockquote><br />
Solving this require us to use F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
to calculate the electric force holding the electron in addition to the gravitational force between the two. <br />
After calculating each, to find the ratio we simply divide the electric force <br />
by the gravitational. Use the proton and electron’s charge for q<SUB>1</SUB> <br />
and q<SUB>2</SUB>, the radius is diameter of a hydrogen atom divided by <br />
two.<br><br />
<br />
Electric Force: F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = (8.99E9)(-1.602E-19)(1.602E-19) / (.53E-10)<SUP>2</SUP> = <br />
8.2E-8 N <br><br />
<br />
Gravitational Force: F = <br />
Gm<SUB>1</SUB>m<SUB>2</SUB>/r<SUP>2</SUP> = <br />
(6.67e-11)(1.673e-27)(9.11e-31) / (.53e-10)<SUP>2</SUP> = <br />
3.6E-47 N <br><br />
<br />
Divide and Conquer: 8.2E-8 / 3.6E-47 = <STRONG>2.3E39 N ratio <br />
Electric/Gravitational</STRONG></P> <br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16.===<br />
:(5.71 x 10<sup>13</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17.===<br />
:(Q<sub>1</sub> = 1/2Q<sub>T</sub> and Q<sub>1</sub>, Q<sub>2</sub> = 0)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
:(2.1x10<sup>-10</sup> m)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===19. A + 5.7 <math>\mu</math>C and a - 3.5 <math>\mu</math>C charge are placed 25 cm apart.; Where can a third charge be placed so that it experiences no net force?===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P19.jpg]]<br />
</div><br />
<blockquote><br />
<P>So we're trying to find r. We want the Forces to be equal, so:</P> <br />
<P>kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>Q<SUB>1</SUB> is the "third charge," so we can just cancel it out here, as <br />
well as the K. So then we're left with:</P> <br />
<P>(3.5 mC)/r<SUP>2</SUP> = (5.7 <br />
mC)/(r+.25)<SUP>2</SUP></P> <br />
<P>Square root top and bottom on both sides...</P> <br />
<P>(3.5 <br />
mC)^(1/2)/r <br />
= (5.7 <br />
mC)^(1/2)/(r+.25)</P> <br />
<P>((3.5 <br />
mC)^(1/2))r <br />
+ ((3.5 <br />
mC)^(1/2))(.25) <br />
= ((5.7 <br />
mC)^(1/2))r</P> <br />
<P>Solving for r, we get:</P> <br />
<P><STRONG>r = .91 m beyond the negative charge</STRONG></P> <br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===20.===<br />
:(50.0x10<sup>-6</sup> C, 30.0x10<sup>-6</sup> C or -15.7x10<sup>-6</sup> C, 95.7x10<sup>-6</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===21.===<br />
<blockquote><br />
The acceleration is produced by the force from the electric field:<br><br />
<br />
F = qE= ma;<br /><br />
<br />
(1.60x10<sup>-19</sup>C)(600N/C) = (9.11x10<sup>-31</sup>kg)a, which gives a= '''1.05x10<sup>14</sup> m/s<sup>2</sup>'''<br /><br />
<br />
Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field <br />
<br />
The direction of the acceleration is independent of the velocity<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
If we take the positive direction to the east, we have <br />
<br />
F = qE = (-1.60x10<sup>-19</sup>C)(+3500N/C) = -5.6x10<sup>-16</sup>N, or '''-5.6x10<sup>-16</sup>N (west)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===23.===<br />
If we take the positive direction to the south, we have<br />
<br />
F = qE;<br />
<br />
3.2x10<sup>-14</sup>N = (+1.60x10<sup>-19</sup>C)E, which gives E = <br />
<br />
'''+2.0x10<sup>5</sup>N/C (south)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
If we take the positive direction up, we have<br />
<br />
F = qE;<br />
<br />
+8.4N = (-8.8x10<sup>-6</sup>C)E, which gives E = <br />
<br />
'''+9.5x10 <sup>5</sup>N/C (up)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===25.===<br />
The electric field above a positive charge will be away from the charge, or up. <br />
<br />
We find the magnitude from <br />
<br />
E = kQ/r<sup>2</sup><br />
<br />
=(9.0x10<sub>9</sub>N*m<sup>2</sup>/C<sup>2</sup>)(33.0x10<sup>-6</sup>C)/(0.020m)<sub>2</sub><br />
<br />
'''=3.30x10<sup>6</sup>N/C (up)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===26.===<br />
The directions of the fields are determined from the signs of the charges are are indicated on the diagram. The net electric field will be to the left. We find its magnitude frm <br />
<br />
E = kQ<sub>1</sub>/L<sup>2</sup> = k(Q<sub>1</sub>+Q<sub>2</sub>)/l<sup>2</sup><br />
<br />
= (9.0x10<sup>9</sup>N*m<sup>2</sup>mC<sup>2</sup>)(8.0x10<sup>-6</sup>C+6.0x10<sup>-6</sup>)/(0.020m)<sup>2</sup><br />
<br />
=3.2x10<sup>8</sup>N/C<br />
<br />
Thus the electric field is '''3.2X10<sup>8</sup>N/C toward the negative charge''' <br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===27.===<br />
The acceleration is produced by the force from the electric field:<br />
<br />
F = qE = ma;<br />
<br />
(-1.60x10<sup>-19</sup>C)E = (9.11x10<sup>-31</sup>kg)(125 m/s<sup>2</sup>), which gives E = -7.12x10<sup>-10</sup>N/C<br />
<br />
Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field, so the electric field is '''-7.12x10<sup>-10</sup>N/C south'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===28.===<br />
The direction of the fields are determined from the signs of the charges and are in the same direction, as indicated on the diagram. The net electric field will be to the left. We find its magnitude from <br />
<br />
E = kQ<sub>1</sub>/L<sup>2</sup>+kQ<sub>2</sub>/L<sup>2</sup> = k(Q=Q)/L<sup>2</sup> = 2kQ/l<sup>2</sup><br />
<br />
1750 N/C = 2(9.0x10<sup>9</sup>N*m<sup>2</sup>/C<sup>2</sup>)Q/(0.020m)<sup>2</sup>, which gives<br />
<br />
'''Q = 6.2x10<sup>-10</sup>C'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 16 41.JPG]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 1.gif ]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 2.gif ]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 3.gif ]]<br />
</div><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===31.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P31.jpg]]<br />
</div><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
:(1.0x10<sup>7</sup> electrons)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
:(A charge of 0.402 Q<sub>o</sub>, 0.366 L from Q<sub>o</sub>)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
==About this page==<br />
All images uploaded for this page must start with the string "Gp5_16_" so 16-38.jpg should be uploaded as Gp5_16_16-38.jpg. This way we can avoid conflicts in the image directory, and we can find images easily.<br><br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_16&diff=298Giancoli Physics (5th ed) Chapter 162007-09-11T22:14:36Z<p>WikiSysop: Reverted edits by WikiSysop (Talk); changed back to last version by Cmurray</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
===1.How many electrons make up a charge of -30.0 <math>\mu</math>C?===<br />
:Since electrons have charges of –1.602 x 10<sup>-19</sup> C all we need to do is divide the given charge by the electron charge to give us the number of electrons. Also, be careful with the whole micro-coulombs thing.<br />
:-30.0 <math>\mu</math>C = -30.0 x 10<sup>-6</sup> C<br />
:(-30.0 x 10<sup>-6</sup> C) /(-1.602 x 10<sup>-19</sup> C) = 1.88 E14 electrons<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(2.7 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. Two charged balls are 20.0 cm apart. They are moved, and the force on each of them is found to have tripled. How far apart are they now?===<br />
:Solving this problem is nice, because in this case we can use the formula <math>F = \frac{kq_1q_2}{r^2}</math> We can set two equations equal to one another. <br />
:<br />
::<math>\frac{3kq_1q_2}{r_1^2} = \frac{kq_1q_2}{r_2^2}</math><br />
:Where r<sub>1</sub> is 0.20 m, and we are trying to solve for r<sub>2</sub><br />
:If we divide both sides by kq<sub>1</sub>q<sub>2</sub> and plug in for r<sub>1</sub> we get:<br />
::<math>\frac{3}{0.20^2} = \frac{1}{r_2^2}</math><br />
:Finally, <br />
:r<sub>2</sub> = .012 m or 11.5 cm<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4.===<br />
:(0.500 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5.What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10<SUP>-12</SUP> m?===<br />
<BLOCKQUOTE> <br />
<P>This is a straight up F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
problem from the hood. <br />
It’s not too bad, the only trick here is the nucleus charge and to solve that <br />
problem we simply multiply the first charge by +26 e. Plug in the values you <br />
must, young Jedi.</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>F = (8.99E9) * (26 * 1.602E-19 * 1.602E-19) / <br />
(1.5E-12)<SUP>2</SUP></P> <br />
<P><STRONG>F = .0027 N or 2.7E-3 N</STRONG></P> <br />
</BLOCKQUOTE><br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6.===<br />
:(9.2 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===7.What is the magnitude of the force a +15 - mC charge exerts on a +3.0-mili coulombs charge 40 cm away? (1 <math>\mu</math>C=10<SUP>-6</SUP> C, 1 mC=10<SUP>-3</SUP>C.)?===<br />
<br />
<BLOCKQUOTE> <br />
<P>Questions like these are ones you cross your fingers for on <br />
the test. Watch out for the whole mili- and mico- coulomb thing and remember <br />
that two positive charges repel. Other than that you can pound this question <br />
out with speed.</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>F = (8.99E9) * (15E-6) * (3E-3) / (.4)<SUP>2</SUP></P> <br />
<P><STRONG>F = 2.5E3 N</STRONG><BR><BR> <br />
<br />
</P><br />
</BLOCKQUOTE><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===8.===<br />
:(3.8x10<sup>14</sup> electrons, 3.4x10<sup>-16</sup> kg)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===9. Imagine space invaders could deposit extra electrons in equal amounts on the earth and now your car, which has mass of 1050 kg. note that the rubber tires would provide some insulation. How much charge Q would need to be placed on your car (same amount on earth) in order to levitate it (overcome gravity) (hint : assume that earth charge is spread uniformly so it acts as if it were located at the earth's center, and then the separation distance is the radius of the earth.)?===<br />
<BLOCKQUOTE> <br />
<P>Ok, seriously man. The author had a little too much time on <br />
his hands to write this problem. Buts it’s assigned so lets get to <br />
it.<BR>Alright, so the first step is to figure out exactly how much force is <br />
required to move your car. Using gravity as the acceleration to counteract, <br />
remember these guys are levitating your car, we can use F = ma.</P> <br />
<P>F = (1050 kg) (9.8 ms-2) = 10,290 N</P> <br />
<p>Now, your car is on the surface of the earth, so the center to center <br />
distance is the radius of the earth essentially, or 6.38 x 10<sup>6</sup> <br />
m.; Setting the coulombic repulsion equal to the force of gravity on the <br />
car we have:<br> <br />
<br> <br />
10,290 N = kq<sub>1</sub>q<sub>2</sub>/r<sup>2<br> <br />
<br> <br />
</sup>Now, since the two charges must be the same (read the problem), this <br />
reduces to:<br> <br />
<br> <br />
10,290 N = kq<sup>2</sup>/r<sup>2<br> <br />
<br> <br />
</sup>10,290 N = (8.99 x 10<sup>9</sup> Nm<sup>2</sup>/C<sup>2</sup>)<b>q<sup>2</sup></b>/(6.38x10<sup>6</sup>m)<sup>2</sup></p> <br />
<P><strong>q = 6825.7 C = <u>6.8e3 C</u></strong><u><BR></u><BR> <br />
<br />
</P></BLOCKQUOTE><br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10.===<br />
:(-5.4x10<sup>7</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. Particles of charge +70, +48, and +80 mC are placed in a line (fig. 16-37) the center one is 0.35m from each of the others. Calculate the net force on each charge due to the other two.===<br />
<div style="text-align: center;"><br />
[[Image:gp5_16_16_37.JPG]]<br />
</div><br><br />
<br />
:<P>First of all, you should have already sketched a diagram of <br />
:the figure because it makes things much, much easier. This problem really <br />
:isn’t that difficult, because we don’t have to deal with angles and instead <br />
:it is <br />
:just plain tedious. Solving this requires you to calculate both forces exerted <br />
:on each particle, determining whether that force is either attractive (like <br />
:me) or repulsive (like my stupid jokes). I will show how to use all the <br />
:necessary calculations for the +70 charge and after that, only the calculated <br />
:values.</P> <br />
:<P>(a) The force on the; +70E-6:;<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP></P> <br />
:<P>+70 and +48:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= <br />
:(8.99E9)(70E-6)(48E-6) / (.35)<SUP>2</SUP> = 246.6 N (left - the 48 repels the 70)</P> <br />
:<P>+70 and –80:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= <br />
:(8.99E9)(70E-6)(-80E-6) / (.7)<SUP>2</SUP> = 102.7 N (right - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:102.7 (right) - 246.6 (left) = <STRONG>-144 N <br />
:(left)</STRONG></P> <br />
:<P>(b) The force on the middle one (The 48E-6 C)<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP></P> <br />
:<P>+48 and +70 :<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(48E-6)(70E-6) / (.35)<SUP>2</SUP> = 246.6 N (right - the 70 repels the 48)</P> <br />
:<P>+48 and –80:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(48E-6)(80E-6) / (.35)<SUP>2</SUP> = 281.8 N (right - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:246.6 N (right) + 281.8 N (right) = <STRONG>528 N (right)</STRONG><BR></P> <br />
:<P>(c) The force on the right one (The -80E-6 C)<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:</P><P>-80 and +48 :<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(80E-6)(48E-6) / (.35)<SUP>2</SUP> = 281.8 N (left - attracted</P> <br />
:<P>-80 and +70:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(80E-6)(70E-6) / (.70)<SUP>2</SUP> = 102.7 N (left - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:281.8 N (left) + 102.7 N (left) = <strong>385 N (left)</strong></P> <br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===12.Three positive particles of charge 11.0 <math>\mu</math>C are located at the corners of an equilateral triangle of side 15.0 cm (Fig. 16-38).; Calculate the magnitude and direction of the net force on each particle.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5_16_16_38.JPG]]<br />
</div><br />
:<P>This problem seems hard because it seems like we’re going to <br />
:end up finding three different answers and plus, they’re not in a straight <br />
:line. However, all we need to do is calculate the magnitude and net force on <br />
:one particle because we are dealing with an equilateral triangle. To solve the <br />
:problem, we need to find the force one particle exerts on the <br />
:other.</P> <br />
:<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = <br />
:(8.99E9)(11E-6)2 / (.15)<SUP>2</SUP> = 48.35 N</P> <br />
:<P>Now, because this is an equilateral triangle, we can assume <br />
:the angle that each particle is exerting on the other is 60<sup>o</sup>. Since the x-axis <br />
:displacement of each particle is the opposite of the other, they cancel one <br />
:another out and all we need to know is the y-axis displacement. To find it we <br />
:use good old Mr. Sine.</P> <br />
:<P>(48.35)(sin 60) = 41.87 * 2 (remember there are two different <br />
:particles acting on each) =</P> <br />
:<P><STRONG>83.7 N (away from the center)</STRONG><BR><BR><br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A charge of 6.00 mC is placed at each corner of a square 1.00 m on a side.; Determine the magnitude and direction of the force on each charge.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P13.jpg]]<br />
</div><br />
<blockquote><br />
Make sure you make a diagram and draw arrows as to where the <br />
force is going. Use F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
and calculate the magnitude and impact <b>each</b> (meaning all 3) charge has <br />
on the other.</P> <br />
<P>First find the force of the two particles that are not on a <br />
diagonal. Since each particle produces and equal force, the charge will be <br />
propelled diagonally, at an angle of 45 degrees.;</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = (8.99E9)(6E-3)<SUP>2</SUP>; <br />
/ (1)<SUP>2</SUP> = 323640 N</P> <br />
<P>323640<SUP>2</SUP> +323640<SUP>2</SUP>; = c<SUP>2</SUP> ; <br />
</P> <br />
<P>c = 457696 N <br />
</P> <br />
<P>Now, we solve for the third charge on the diagonal. <br />
</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = <br />
(8.99E9)(6E-3)<SUP>2</SUP>; / 2 <br />
</P> <br />
<P>F = 161820 N <br />
</P><P>Now just add the two and the answer is yours. <br />
</P><P>F = 161820 + 457696 = 619516 N = <b>6.2 E5 N</b> <br />
</P> <br />
<P><br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===14.===<br />
2.96x10<sup>5</sup> N Toward the center of the square<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15.Compare the electric force holding the electron in orbit around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton.; What is the ratio of the these two forces?===<br />
<br />
<blockquote><br />
Solving this require us to use F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
to calculate the electric force holding the electron in addition to the gravitational force between the two. <br />
After calculating each, to find the ratio we simply divide the electric force <br />
by the gravitational. Use the proton and electron’s charge for q<SUB>1</SUB> <br />
and q<SUB>2</SUB>, the radius is diameter of a hydrogen atom divided by <br />
two.<br><br />
<br />
Electric Force: F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = (8.99E9)(-1.602E-19)(1.602E-19) / (.53E-10)<SUP>2</SUP> = <br />
8.2E-8 N <br><br />
<br />
Gravitational Force: F = <br />
Gm<SUB>1</SUB>m<SUB>2</SUB>/r<SUP>2</SUP> = <br />
(6.67e-11)(1.673e-27)(9.11e-31) / (.53e-10)<SUP>2</SUP> = <br />
3.6E-47 N <br><br />
<br />
Divide and Conquer: 8.2E-8 / 3.6E-47 = <STRONG>2.3E39 N ratio <br />
Electric/Gravitational</STRONG></P> <br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16.===<br />
:(5.71 x 10<sup>13</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17.===<br />
:(Q<sub>1</sub> = 1/2Q<sub>T</sub> and Q<sub>1</sub>, Q<sub>2</sub> = 0)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
:(2.1x10<sup>-10</sup> m)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===19. A + 5.7 <math>\mu</math>C and a - 3.5 <math>\mu</math>C charge are placed 25 cm apart.; Where can a third charge be placed so that it experiences no net force?===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P19.jpg]]<br />
</div><br />
<blockquote><br />
<P>So we're trying to find r. We want the Forces to be equal, so:</P> <br />
<P>kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>Q<SUB>1</SUB> is the "third charge," so we can just cancel it out here, as <br />
well as the K. So then we're left with:</P> <br />
<P>(3.5 mC)/r<SUP>2</SUP> = (5.7 <br />
mC)/(r+.25)<SUP>2</SUP></P> <br />
<P>Square root top and bottom on both sides...</P> <br />
<P>(3.5 <br />
mC)^(1/2)/r <br />
= (5.7 <br />
mC)^(1/2)/(r+.25)</P> <br />
<P>((3.5 <br />
mC)^(1/2))r <br />
+ ((3.5 <br />
mC)^(1/2))(.25) <br />
= ((5.7 <br />
mC)^(1/2))r</P> <br />
<P>Solving for r, we get:</P> <br />
<P><STRONG>r = .91 m beyond the negative charge</STRONG></P> <br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===20.===<br />
:(50.0x10<sup>-6</sup> C, 30.0x10<sup>-6</sup> C or -15.7x10<sup>-6</sup> C, 95.7x10<sup>-6</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===21.===<br />
<blockquote><br />
The acceleration is produced by the force from the electric field:<br><br />
<br />
F = qE= ma;<br /><br />
<br />
(1.60x10<sup>-19</sup>C)(600N/C) = (9.11x10<sup>-31</sup>kg)a, which gives a= '''1.05x10<sup>14</sup> m/s<sup>2</sup>'''<br /><br />
<br />
Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field <br />
<br />
The direction of the acceleration is independent of the velocity<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
If we take the positive direction to the east, we have <br />
<br />
F = qE = (-1.60x10<sup>-19</sup>C)(+3500N/C) = -5.6x10<sup>-16</sup>N, or '''-5.6x10<sup>-16</sup>N (west)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===23.===<br />
If we take the positive direction to the south, we have<br />
<br />
F = qE;<br />
<br />
3.2x10<sup>-14</sup>N = (+1.60x10<sup>-19</sup>C)E, which gives E = <br />
<br />
'''+2.0x10<sup>5</sup>N/C (south)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
If we take the positive direction up, we have<br />
<br />
F = qE;<br />
<br />
+8.4N = (-8.8x10<sup>-6</sup>C)E, which gives E = <br />
<br />
'''+9.5x10 <sup>5</sup>N/C (up)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===25.===<br />
The electric field above a positive charge will be away from the charge, or up. <br />
<br />
We find the magnitude from <br />
<br />
E = kQ/r<sup>2</sup><br />
<br />
=(9.0x10<sub>9</sub>N*m<sup>2</sup>/C<sup>2</sup>)(33.0x10<sup>-6</sup>C)/(0.020m)<sub>2</sub><br />
<br />
'''=3.30x10<sup>6</sup>N/C (up)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===26.===<br />
The directions of the fields are determined from the signs of the charges are are indicated on the diagram. The net electric field will be to the left. We find its magnitude frm <br />
<br />
E = kQ<sub>1</sub>/L<sup>2</sup> = k(Q<sub>1</sub>+Q<sub>2</sub>)/l<sup>2</sup><br />
<br />
= (9.0x10<sup>9</sup>N*m<sup>2</sup>mC<sup>2</sup>)(8.0x10<sup>-6</sup>C+6.0x10<sup>-6</sup>)/(0.020m)<sup>2</sup><br />
<br />
=3.2x10<sup>8</sup>N/C<br />
<br />
Thus the electric field is '''3.2X10<sup>8</sup>N/C toward the negative charge''' <br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===27.===<br />
The acceleration is produced by the force from the electric field:<br />
<br />
F = qE = ma;<br />
<br />
(-1.60x10<sup>-19</sup>C)E = (9.11x10<sup>-31</sup>kg)(125 m/s<sup>2</sup>), which gives E = -7.12x10<sup>-10</sup>N/C<br />
<br />
Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field, so the electric field is '''-7.12x10<sup>-10</sup>N/C south'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===28.===<br />
The direction of the fields are determined from the signs of the charges and are in the same direction, as indicated on the diagram. The net electric field will be to the left. We find its magnitude from <br />
<br />
E = kQ<sub>1</sub>/L<sup>2</sup>+kQ<sub>2</sub>/L<sup>2</sup> = k(Q=Q)/L<sup>2</sup> = 2kQ/l<sup>2</sup><br />
<br />
1750 N/C = 2(9.0x10<sup>9</sup>N*m<sup>2</sup>/C<sup>2</sup>)Q/(0.020m)<sup>2</sup>, which gives<br />
<br />
'''Q = 6.2x10<sup>-10</sup>C'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 16 41.JPG]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 1.gif ]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 2.gif ]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 3.gif ]]<br />
</div><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===31.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P31.jpg]]<br />
</div><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
:(1.0x10<sup>7</sup> electrons)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
:(A charge of 0.402 Q<sub>o</sub>, 0.366 L from Q<sub>o</sub>)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
==About this page==<br />
All images uploaded for this page must start with the string "Gp5_16_" so 16-38.jpg should be uploaded as Gp5_16_16-38.jpg. This way we can avoid conflicts in the image directory, and we can find images easily.<br><br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_16&diff=297Giancoli Physics (5th ed) Chapter 162007-09-11T22:14:09Z<p>WikiSysop: Undo revision 288 by Special:Contributions/Cnishitani (User talk:Cnishitani)</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
===1.How many electrons make up a charge of -30.0 <math>\mu</math>C?===<br />
:Since electrons have charges of –1.602 x 10<sup>-19</sup> C all we need to do is divide the given charge by the electron charge to give us the number of electrons. Also, be careful with the whole micro-coulombs thing.<br />
:-30.0 <math>\mu</math>C = -30.0 x 10<sup>-6</sup> C<br />
:(-30.0 x 10<sup>-6</sup> C) /(-1.602 x 10<sup>-19</sup> C) = 1.88 E14 electrons<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(2.7 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. Two charged balls are 20.0 cm apart. They are moved, and the force on each of them is found to have tripled. How far apart are they now?===<br />
:Solving this problem is nice, because in this case we can use the formula <math>F = \frac{kq_1q_2}{r^2}</math> We can set two equations equal to one another. <br />
:<br />
::<math>\frac{3kq_1q_2}{r_1^2} = \frac{kq_1q_2}{r_2^2}</math><br />
:Where r<sub>1</sub> is 0.20 m, and we are trying to solve for r<sub>2</sub><br />
:If we divide both sides by kq<sub>1</sub>q<sub>2</sub> and plug in for r<sub>1</sub> we get:<br />
::<math>\frac{3}{0.20^2} = \frac{1}{r_2^2}</math><br />
:Finally, <br />
:r<sub>2</sub> = .012 m or 11.5 cm<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4.===<br />
:(0.500 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5.What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10<SUP>-12</SUP> m?===<br />
<BLOCKQUOTE> <br />
<P>This is a straight up F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
problem from the hood. <br />
It’s not too bad, the only trick here is the nucleus charge and to solve that <br />
problem we simply multiply the first charge by +26 e. Plug in the values you <br />
must, young Jedi.</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>F = (8.99E9) * (26 * 1.602E-19 * 1.602E-19) / <br />
(1.5E-12)<SUP>2</SUP></P> <br />
<P><STRONG>F = .0027 N or 2.7E-3 N</STRONG></P> <br />
</BLOCKQUOTE><br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6.===<br />
:(9.2 N)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===7.What is the magnitude of the force a +15 - mC charge exerts on a +3.0-mili coulombs charge 40 cm away? (1 <math>\mu</math>C=10<SUP>-6</SUP> C, 1 mC=10<SUP>-3</SUP>C.)?===<br />
<br />
<BLOCKQUOTE> <br />
<P>Questions like these are ones you cross your fingers for on <br />
the test. Watch out for the whole mili- and mico- coulomb thing and remember <br />
that two positive charges repel. Other than that you can pound this question <br />
out with speed.</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>F = (8.99E9) * (15E-6) * (3E-3) / (.4)<SUP>2</SUP></P> <br />
<P><STRONG>F = 2.5E3 N</STRONG><BR><BR> <br />
<br />
</P><br />
</BLOCKQUOTE><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===8.===<br />
:(3.8x10<sup>14</sup> electrons, 3.4x10<sup>-16</sup> kg)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===9. Imagine space invaders could deposit extra electrons in equal amounts on the earth and now your car, which has mass of 1050 kg. note that the rubber tires would provide some insulation. How much charge Q would need to be placed on your car (same amount on earth) in order to levitate it (overcome gravity) (hint : assume that earth charge is spread uniformly so it acts as if it were located at the earth's center, and then the separation distance is the radius of the earth.)?===<br />
<BLOCKQUOTE> <br />
<P>Ok, seriously man. The author had a little too much time on <br />
his hands to write this problem. Buts it’s assigned so lets get to <br />
it.<BR>Alright, so the first step is to figure out exactly how much force is <br />
required to move your car. Using gravity as the acceleration to counteract, <br />
remember these guys are levitating your car, we can use F = ma.</P> <br />
<P>F = (1050 kg) (9.8 ms-2) = 10,290 N</P> <br />
<p>Now, your car is on the surface of the earth, so the center to center <br />
distance is the radius of the earth essentially, or 6.38 x 10<sup>6</sup> <br />
m.; Setting the coulombic repulsion equal to the force of gravity on the <br />
car we have:<br> <br />
<br> <br />
10,290 N = kq<sub>1</sub>q<sub>2</sub>/r<sup>2<br> <br />
<br> <br />
</sup>Now, since the two charges must be the same (read the problem), this <br />
reduces to:<br> <br />
<br> <br />
10,290 N = kq<sup>2</sup>/r<sup>2<br> <br />
<br> <br />
</sup>10,290 N = (8.99 x 10<sup>9</sup> Nm<sup>2</sup>/C<sup>2</sup>)<b>q<sup>2</sup></b>/(6.38x10<sup>6</sup>m)<sup>2</sup></p> <br />
<P><strong>q = 6825.7 C = <u>6.8e3 C</u></strong><u><BR></u><BR> <br />
<br />
</P></BLOCKQUOTE><br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10.===<br />
:(-5.4x10<sup>7</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. Particles of charge +70, +48, and +80 mC are placed in a line (fig. 16-37) the center one is 0.35m from each of the others. Calculate the net force on each charge due to the other two.===<br />
<div style="text-align: center;"><br />
[[Image:gp5_16_16_37.JPG]]<br />
</div><br><br />
<br />
:<P>First of all, you should have already sketched a diagram of <br />
:the figure because it makes things much, much easier. This problem really <br />
:isn’t that difficult, because we don’t have to deal with angles and instead <br />
:it is <br />
:just plain tedious. Solving this requires you to calculate both forces exerted <br />
:on each particle, determining whether that force is either attractive (like <br />
:me) or repulsive (like my stupid jokes). I will show how to use all the <br />
:necessary calculations for the +70 charge and after that, only the calculated <br />
:values.</P> <br />
:<P>(a) The force on the; +70E-6:;<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP></P> <br />
:<P>+70 and +48:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= <br />
:(8.99E9)(70E-6)(48E-6) / (.35)<SUP>2</SUP> = 246.6 N (left - the 48 repels the 70)</P> <br />
:<P>+70 and –80:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= <br />
:(8.99E9)(70E-6)(-80E-6) / (.7)<SUP>2</SUP> = 102.7 N (right - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:102.7 (right) - 246.6 (left) = <STRONG>-144 N <br />
:(left)</STRONG></P> <br />
:<P>(b) The force on the middle one (The 48E-6 C)<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP></P> <br />
:<P>+48 and +70 :<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(48E-6)(70E-6) / (.35)<SUP>2</SUP> = 246.6 N (right - the 70 repels the 48)</P> <br />
:<P>+48 and –80:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(48E-6)(80E-6) / (.35)<SUP>2</SUP> = 281.8 N (right - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:246.6 N (right) + 281.8 N (right) = <STRONG>528 N (right)</STRONG><BR></P> <br />
:<P>(c) The force on the right one (The -80E-6 C)<br> <br />
:Using F = <br />
:KQ<SUB>1</SUB>Q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:</P><P>-80 and +48 :<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(80E-6)(48E-6) / (.35)<SUP>2</SUP> = 281.8 N (left - attracted</P> <br />
:<P>-80 and +70:<br> <br />
:F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
:= (8.99E9)(80E-6)(70E-6) / (.70)<SUP>2</SUP> = 102.7 N (left - attracted)</P> <br />
:<P>Then, just do a little net force action:<br> <br />
:281.8 N (left) + 102.7 N (left) = <strong>385 N (left)</strong></P> <br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===12.Three positive particles of charge 11.0 <math>\mu</math>C are located at the corners of an equilateral triangle of side 15.0 cm (Fig. 16-38).; Calculate the magnitude and direction of the net force on each particle.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5_16_16_38.JPG]]<br />
</div><br />
:<P>This problem seems hard because it seems like we’re going to <br />
:end up finding three different answers and plus, they’re not in a straight <br />
:line. However, all we need to do is calculate the magnitude and net force on <br />
:one particle because we are dealing with an equilateral triangle. To solve the <br />
:problem, we need to find the force one particle exerts on the <br />
:other.</P> <br />
:<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = <br />
:(8.99E9)(11E-6)2 / (.15)<SUP>2</SUP> = 48.35 N</P> <br />
:<P>Now, because this is an equilateral triangle, we can assume <br />
:the angle that each particle is exerting on the other is 60<sup>o</sup>. Since the x-axis <br />
:displacement of each particle is the opposite of the other, they cancel one <br />
:another out and all we need to know is the y-axis displacement. To find it we <br />
:use good old Mr. Sine.</P> <br />
:<P>(48.35)(sin 60) = 41.87 * 2 (remember there are two different <br />
:particles acting on each) =</P> <br />
:<P><STRONG>83.7 N (away from the center)</STRONG><BR><BR><br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A charge of 6.00 mC is placed at each corner of a square 1.00 m on a side.; Determine the magnitude and direction of the force on each charge.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P13.jpg]]<br />
</div><br />
<blockquote><br />
Make sure you make a diagram and draw arrows as to where the <br />
force is going. Use F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
and calculate the magnitude and impact <b>each</b> (meaning all 3) charge has <br />
on the other.</P> <br />
<P>First find the force of the two particles that are not on a <br />
diagonal. Since each particle produces and equal force, the charge will be <br />
propelled diagonally, at an angle of 45 degrees.;</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = (8.99E9)(6E-3)<SUP>2</SUP>; <br />
/ (1)<SUP>2</SUP> = 323640 N</P> <br />
<P>323640<SUP>2</SUP> +323640<SUP>2</SUP>; = c<SUP>2</SUP> ; <br />
</P> <br />
<P>c = 457696 N <br />
</P> <br />
<P>Now, we solve for the third charge on the diagonal. <br />
</P> <br />
<P>F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = <br />
(8.99E9)(6E-3)<SUP>2</SUP>; / 2 <br />
</P> <br />
<P>F = 161820 N <br />
</P><P>Now just add the two and the answer is yours. <br />
</P><P>F = 161820 + 457696 = 619516 N = <b>6.2 E5 N</b> <br />
</P> <br />
<P><br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===14.===<br />
2.96x10<sup>5</sup> N Toward the center of the square<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15.Compare the electric force holding the electron in orbit around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton.; What is the ratio of the these two forces?===<br />
<br />
<blockquote><br />
Solving this require us to use F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> <br />
to calculate the electric force holding the electron in addition to the gravitational force between the two. <br />
After calculating each, to find the ratio we simply divide the electric force <br />
by the gravitational. Use the proton and electron’s charge for q<SUB>1</SUB> <br />
and q<SUB>2</SUB>, the radius is diameter of a hydrogen atom divided by <br />
two.<br><br />
<br />
Electric Force: F = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = (8.99E9)(-1.602E-19)(1.602E-19) / (.53E-10)<SUP>2</SUP> = <br />
8.2E-8 N <br><br />
<br />
Gravitational Force: F = <br />
Gm<SUB>1</SUB>m<SUB>2</SUB>/r<SUP>2</SUP> = <br />
(6.67e-11)(1.673e-27)(9.11e-31) / (.53e-10)<SUP>2</SUP> = <br />
3.6E-47 N <br><br />
<br />
Divide and Conquer: 8.2E-8 / 3.6E-47 = <STRONG>2.3E39 N ratio <br />
Electric/Gravitational</STRONG></P> <br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16.===<br />
:(5.71 x 10<sup>13</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17.===<br />
:(Q<sub>1</sub> = 1/2Q<sub>T</sub> and Q<sub>1</sub>, Q<sub>2</sub> = 0)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
:(2.1x10<sup>-10</sup> m)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===19. A + 5.7 <math>\mu</math>C and a - 3.5 <math>\mu</math>C charge are placed 25 cm apart.; Where can a third charge be placed so that it experiences no net force?===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P19.jpg]]<br />
</div><br />
<blockquote><br />
<P>So we're trying to find r. We want the Forces to be equal, so:</P> <br />
<P>kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> = kq<SUB>1</SUB>q<SUB>2</SUB>/r<SUP>2</SUP> </P> <br />
<P>Q<SUB>1</SUB> is the "third charge," so we can just cancel it out here, as <br />
well as the K. So then we're left with:</P> <br />
<P>(3.5 mC)/r<SUP>2</SUP> = (5.7 <br />
mC)/(r+.25)<SUP>2</SUP></P> <br />
<P>Square root top and bottom on both sides...</P> <br />
<P>(3.5 <br />
mC)^(1/2)/r <br />
= (5.7 <br />
mC)^(1/2)/(r+.25)</P> <br />
<P>((3.5 <br />
mC)^(1/2))r <br />
+ ((3.5 <br />
mC)^(1/2))(.25) <br />
= ((5.7 <br />
mC)^(1/2))r</P> <br />
<P>Solving for r, we get:</P> <br />
<P><STRONG>r = .91 m beyond the negative charge</STRONG></P> <br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===20.===<br />
:(50.0x10<sup>-6</sup> C, 30.0x10<sup>-6</sup> C or -15.7x10<sup>-6</sup> C, 95.7x10<sup>-6</sup> C)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===21.===<br />
<blockquote><br />
The acceleration is produced by the force from the electric field:<br><br />
<br />
F = qE= ma;<br /><br />
<br />
(1.60x10<sup>-19</sup>C)(600N/C) = (9.11x10<sup>-31</sup>kg)a, which gives a= '''1.05x10<sup>14</sup> m/s<sup>2</sup>'''<br /><br />
<br />
Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field <br />
<br />
The direction of the acceleration is independent of the velocity<br />
</blockquote><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
If we take the positive direction to the south, we have<br />
<br />
F = qE;<br />
<br />
3.2x10<sup>-14</sup>N = (+1.60x10<sup>-19</sup>C)E, which gives E = <br />
<br />
'''+2.0x10<sup>5</sup>N/C (south)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
If we take the positive direction up, we have<br />
<br />
F = qE;<br />
<br />
+8.4N = (-8.8x10<sup>-6</sup>C)E, which gives E = <br />
<br />
'''+9.5x10 <sup>5</sup>N/C (up)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===25.===<br />
The electric field above a positive charge will be away from the charge, or up. <br />
<br />
We find the magnitude from <br />
<br />
E = kQ/r<sup>2</sup><br />
<br />
=(9.0x10<sub>9</sub>N*m<sup>2</sup>/C<sup>2</sup>)(33.0x10<sup>-6</sup>C)/(0.020m)<sub>2</sub><br />
<br />
'''=3.30x10<sup>6</sup>N/C (up)'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===26.===<br />
The directions of the fields are determined from the signs of the charges are are indicated on the diagram. The net electric field will be to the left. We find its magnitude frm <br />
<br />
E = kQ<sub>1</sub>/L<sup>2</sup> = k(Q<sub>1</sub>+Q<sub>2</sub>)/l<sup>2</sup><br />
<br />
= (9.0x10<sup>9</sup>N*m<sup>2</sup>mC<sup>2</sup>)(8.0x10<sup>-6</sup>C+6.0x10<sup>-6</sup>)/(0.020m)<sup>2</sup><br />
<br />
=3.2x10<sup>8</sup>N/C<br />
<br />
Thus the electric field is '''3.2X10<sup>8</sup>N/C toward the negative charge''' <br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===27.===<br />
The acceleration is produced by the force from the electric field:<br />
<br />
F = qE = ma;<br />
<br />
(-1.60x10<sup>-19</sup>C)E = (9.11x10<sup>-31</sup>kg)(125 m/s<sup>2</sup>), which gives E = -7.12x10<sup>-10</sup>N/C<br />
<br />
Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field, so the electric field is '''-7.12x10<sup>-10</sup>N/C south'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===28.===<br />
The direction of the fields are determined from the signs of the charges and are in the same direction, as indicated on the diagram. The net electric field will be to the left. We find its magnitude from <br />
<br />
E = kQ<sub>1</sub>/L<sup>2</sup>+kQ<sub>2</sub>/L<sup>2</sup> = k(Q=Q)/L<sup>2</sup> = 2kQ/l<sup>2</sup><br />
<br />
1750 N/C = 2(9.0x10<sup>9</sup>N*m<sup>2</sup>/C<sup>2</sup>)Q/(0.020m)<sup>2</sup>, which gives<br />
<br />
'''Q = 6.2x10<sup>-10</sup>C'''<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 16 41.JPG]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 1.gif ]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 2.gif ]]<br />
</div><br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 number30 3.gif ]]<br />
</div><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===31.===<br />
<div style="text-align: center;"><br />
[[Image:Gp5 16 P31.jpg]]<br />
</div><br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
:(1.0x10<sup>7</sup> electrons)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
:(A charge of 0.402 Q<sub>o</sub>, 0.366 L from Q<sub>o</sub>)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
==About this page==<br />
All images uploaded for this page must start with the string "Gp5_16_" so 16-38.jpg should be uploaded as Gp5_16_16-38.jpg. This way we can avoid conflicts in the image directory, and we can find images easily.<br><br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=118Giancoli Physics (5th ed) Chapter 22007-09-05T18:06:27Z<p>WikiSysop: /* 27. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.A car traveling 45 km/h slows down at a constant .50 m/s/s just by "letting up on the gas". Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop and (c) the distance it travels during the first and fifth seconds.===<br />
-Change 45 km/h into m/s<br />
<br />
45 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 12.5 m/s<br />
<br />
-Remember that slowing down is a negative acceleration. So, a = - .5 m/s/s<br />
<br />
-(a) The initial velocity (u) is 12.5 m/s, the acceleration is -.5 m/s/s, and the final velocity (v) is 0 m/s, because we want it to stop. Using the formula v2 = u2 + 2as, calculate the distance it takes to stop.<br />
<br />
02 = (12.5)2 +2(-.5)s<br />
<br />
s = 156.25 m or 1.6 x 102 m<br />
<br />
-(b) To find the time it takes to stop, put the information in to the formula: v = u + at.<br />
<br />
0 = 12.5 + (-.5)t<br />
<br />
t = 25 s<br />
<br />
-(c) The distance the car traveled during the first second can be found by using the formula s = ut + 1/2 at2, and substituting 1 s in for t.<br />
<br />
s = (12.5)(1) + (.5)(-.5)(1)2<br />
<br />
s = 12.25 m or 12 m<br />
<br />
-During the fifth second, you have to realize that the distance traveled during this time is the distance between the fifth and sixth seconds. So to find the distance during only the fifth second, you take the distance traveled up to the sixth second, and subtract the distance traveled up to the fifth second.<br />
<br />
s = (12.5)(6) + (.5)(-.5)(6)2 s = 66 m<br />
<br />
s = (12.5)(5) + (.5)(-.5)(5)2 s = 56.25 m<br />
<br />
Now take the distance up to the sixth second and subtract the distance up to the fifth.<br />
<br />
66 m - 56.25 m = 9.75 m or 10 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27. Determine the stopping distances for an automobile with an initial speed of 90 km/h and a human reaction time of 1.0s: (a) for an acceleration a= - 4.0 m/s/s; (b) for an acceleration a= - 8 m/s/s.===<br />
-Change 90 km/h into m/s.<br />
<br />
:90 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 25 m/s<br />
<br />
:-(a) First, find the distance it takes to stop with an acceleration of - 4.0 m/s/s. Use the formula: v2 = u2 + 2as<br />
<br />
:02 = 252 + 2(-4.0)s<br />
<br />
:s = 78.125 m<br />
<br />
:-But remember that it takes one second for human reaction, so before the driver starts slowing down, the car is traveling at 25 m/s for one second. The car travels a distance of 25 m.<br />
<br />
:-Add 25 m to the 78.125 m, and you get 103 m.<br />
<br />
:-(b) Do the same as with a, but for an acceleration of -8.0 m/s/s.<br />
<br />
:02 = 252 + 2(-8.0)s<br />
<br />
:s = 39.06 m<br />
<br />
:-Add 39.06 m to the 25 m the car travels during the one second of human reaction to get 64 m.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
-The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at<br />
<br />
0 = 5.5 + (-9.8)t<br />
<br />
-5.5 = (-9.8)t<br />
<br />
t = .56 s<br />
<br />
-Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.<br />
<br />
02 = (5.5)2 +2(-9.8)s<br />
<br />
s = 1.54 m<br />
<br />
-Add this to the 105 m, and we get 106 .54 m.<br />
<br />
-How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2<br />
<br />
106.54 = (0)t + 1/2(9.8)t2<br />
<br />
t = 4.66 s<br />
<br />
-Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?===<br />
-We can use the formula s = ut + 1/2 at2, but the initial velocity (u) is zero, so we are left with, s = 1/2 at2. Replacing h with s, and solving for time gives us tf (the time to fall) = (2h)/g)^1/2<br />
<br />
-The time it takes for the sound to be heard up at the top of the cliff can be found by the formula v = s/t or t = s/v, which is ts = h /(340 m/s).<br />
<br />
-The total time is 3.4 s, so ts + tf = 3.4 s<br />
<br />
-Substitute for ts and tf. This gives us:<br />
<br />
((h /340) + (2h /g)^1/2 = 3.4)<br />
<br />
-This is the same as:<br />
<br />
(1/340) ((h)^1/2)2 + (2/g)^1/2 (h)^1/2 - 3.4 = 0<br />
<br />
-This is quadratic now, where x = h^1/2, so solve it using the quadratic formula (in your book on page 1042).<br />
<br />
h = 52 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?===<br />
:-(a) If it is thrown upward, we must find the time it takes to stop going upward, and reach a final velocity (v) of zero. Using the formula v = u + at, find the time in the air.<br />
<br />
:0 = 12 + (-9.8)t<br />
<br />
:t = 1.23 s<br />
<br />
:-Now find the height it went during that 1.22 s using the formula v2 = u2 + 2as.<br />
<br />
:02 = (12)2 + 2(-9.8)s<br />
<br />
:s = 7.35 m<br />
<br />
:-Add 7.35 m to the height of the cliff, 75 m, to get 82.35 m. Now find the time it takes to fall this new height using the formula s = ut + 1/2 at2<br />
<br />
:82.35 m = (0)t + 1/2(-9.8)t2<br />
<br />
:t = 4.1 s<br />
<br />
:-Add 4.1 s to the time the stone is going upward, which is 1.23 s, and you get 5.33 s.<br />
<br />
:-(b) The initial velocity (u) is 0 m/s and the height is 82.35 m. Using the formula v2 = u2 + 2as, find the final velocity.<br />
<br />
:v2 = (0)2 + 2(-9.8)(82.35)<br />
<br />
:v = - 40.2 m/s (negative because it is going down)<br />
<br />
:-(c) The total distance it traveled would be the distance it traveled going up, 7.35 m, and the total distance it went to the bottom of the cliff, 82.35 m. So adding these together, we get, 89.7 m.<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=117Giancoli Physics (5th ed) Chapter 22007-09-05T18:04:39Z<p>WikiSysop: /* 25. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.A car traveling 45 km/h slows down at a constant .50 m/s/s just by "letting up on the gas". Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop and (c) the distance it travels during the first and fifth seconds.===<br />
-Change 45 km/h into m/s<br />
<br />
45 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 12.5 m/s<br />
<br />
-Remember that slowing down is a negative acceleration. So, a = - .5 m/s/s<br />
<br />
-(a) The initial velocity (u) is 12.5 m/s, the acceleration is -.5 m/s/s, and the final velocity (v) is 0 m/s, because we want it to stop. Using the formula v2 = u2 + 2as, calculate the distance it takes to stop.<br />
<br />
02 = (12.5)2 +2(-.5)s<br />
<br />
s = 156.25 m or 1.6 x 102 m<br />
<br />
-(b) To find the time it takes to stop, put the information in to the formula: v = u + at.<br />
<br />
0 = 12.5 + (-.5)t<br />
<br />
t = 25 s<br />
<br />
-(c) The distance the car traveled during the first second can be found by using the formula s = ut + 1/2 at2, and substituting 1 s in for t.<br />
<br />
s = (12.5)(1) + (.5)(-.5)(1)2<br />
<br />
s = 12.25 m or 12 m<br />
<br />
-During the fifth second, you have to realize that the distance traveled during this time is the distance between the fifth and sixth seconds. So to find the distance during only the fifth second, you take the distance traveled up to the sixth second, and subtract the distance traveled up to the fifth second.<br />
<br />
s = (12.5)(6) + (.5)(-.5)(6)2 s = 66 m<br />
<br />
s = (12.5)(5) + (.5)(-.5)(5)2 s = 56.25 m<br />
<br />
Now take the distance up to the sixth second and subtract the distance up to the fifth.<br />
<br />
66 m - 56.25 m = 9.75 m or 10 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
-The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at<br />
<br />
0 = 5.5 + (-9.8)t<br />
<br />
-5.5 = (-9.8)t<br />
<br />
t = .56 s<br />
<br />
-Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.<br />
<br />
02 = (5.5)2 +2(-9.8)s<br />
<br />
s = 1.54 m<br />
<br />
-Add this to the 105 m, and we get 106 .54 m.<br />
<br />
-How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2<br />
<br />
106.54 = (0)t + 1/2(9.8)t2<br />
<br />
t = 4.66 s<br />
<br />
-Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?===<br />
-We can use the formula s = ut + 1/2 at2, but the initial velocity (u) is zero, so we are left with, s = 1/2 at2. Replacing h with s, and solving for time gives us tf (the time to fall) = (2h)/g)^1/2<br />
<br />
-The time it takes for the sound to be heard up at the top of the cliff can be found by the formula v = s/t or t = s/v, which is ts = h /(340 m/s).<br />
<br />
-The total time is 3.4 s, so ts + tf = 3.4 s<br />
<br />
-Substitute for ts and tf. This gives us:<br />
<br />
((h /340) + (2h /g)^1/2 = 3.4)<br />
<br />
-This is the same as:<br />
<br />
(1/340) ((h)^1/2)2 + (2/g)^1/2 (h)^1/2 - 3.4 = 0<br />
<br />
-This is quadratic now, where x = h^1/2, so solve it using the quadratic formula (in your book on page 1042).<br />
<br />
h = 52 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?===<br />
:-(a) If it is thrown upward, we must find the time it takes to stop going upward, and reach a final velocity (v) of zero. Using the formula v = u + at, find the time in the air.<br />
<br />
:0 = 12 + (-9.8)t<br />
<br />
:t = 1.23 s<br />
<br />
:-Now find the height it went during that 1.22 s using the formula v2 = u2 + 2as.<br />
<br />
:02 = (12)2 + 2(-9.8)s<br />
<br />
:s = 7.35 m<br />
<br />
:-Add 7.35 m to the height of the cliff, 75 m, to get 82.35 m. Now find the time it takes to fall this new height using the formula s = ut + 1/2 at2<br />
<br />
:82.35 m = (0)t + 1/2(-9.8)t2<br />
<br />
:t = 4.1 s<br />
<br />
:-Add 4.1 s to the time the stone is going upward, which is 1.23 s, and you get 5.33 s.<br />
<br />
:-(b) The initial velocity (u) is 0 m/s and the height is 82.35 m. Using the formula v2 = u2 + 2as, find the final velocity.<br />
<br />
:v2 = (0)2 + 2(-9.8)(82.35)<br />
<br />
:v = - 40.2 m/s (negative because it is going down)<br />
<br />
:-(c) The total distance it traveled would be the distance it traveled going up, 7.35 m, and the total distance it went to the bottom of the cliff, 82.35 m. So adding these together, we get, 89.7 m.<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=116Giancoli Physics (5th ed) Chapter 22007-09-05T18:03:11Z<p>WikiSysop: /* 49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed jus</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
-The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at<br />
<br />
0 = 5.5 + (-9.8)t<br />
<br />
-5.5 = (-9.8)t<br />
<br />
t = .56 s<br />
<br />
-Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.<br />
<br />
02 = (5.5)2 +2(-9.8)s<br />
<br />
s = 1.54 m<br />
<br />
-Add this to the 105 m, and we get 106 .54 m.<br />
<br />
-How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2<br />
<br />
106.54 = (0)t + 1/2(9.8)t2<br />
<br />
t = 4.66 s<br />
<br />
-Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?===<br />
-We can use the formula s = ut + 1/2 at2, but the initial velocity (u) is zero, so we are left with, s = 1/2 at2. Replacing h with s, and solving for time gives us tf (the time to fall) = (2h)/g)^1/2<br />
<br />
-The time it takes for the sound to be heard up at the top of the cliff can be found by the formula v = s/t or t = s/v, which is ts = h /(340 m/s).<br />
<br />
-The total time is 3.4 s, so ts + tf = 3.4 s<br />
<br />
-Substitute for ts and tf. This gives us:<br />
<br />
((h /340) + (2h /g)^1/2 = 3.4)<br />
<br />
-This is the same as:<br />
<br />
(1/340) ((h)^1/2)2 + (2/g)^1/2 (h)^1/2 - 3.4 = 0<br />
<br />
-This is quadratic now, where x = h^1/2, so solve it using the quadratic formula (in your book on page 1042).<br />
<br />
h = 52 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?===<br />
:-(a) If it is thrown upward, we must find the time it takes to stop going upward, and reach a final velocity (v) of zero. Using the formula v = u + at, find the time in the air.<br />
<br />
:0 = 12 + (-9.8)t<br />
<br />
:t = 1.23 s<br />
<br />
:-Now find the height it went during that 1.22 s using the formula v2 = u2 + 2as.<br />
<br />
:02 = (12)2 + 2(-9.8)s<br />
<br />
:s = 7.35 m<br />
<br />
:-Add 7.35 m to the height of the cliff, 75 m, to get 82.35 m. Now find the time it takes to fall this new height using the formula s = ut + 1/2 at2<br />
<br />
:82.35 m = (0)t + 1/2(-9.8)t2<br />
<br />
:t = 4.1 s<br />
<br />
:-Add 4.1 s to the time the stone is going upward, which is 1.23 s, and you get 5.33 s.<br />
<br />
:-(b) The initial velocity (u) is 0 m/s and the height is 82.35 m. Using the formula v2 = u2 + 2as, find the final velocity.<br />
<br />
:v2 = (0)2 + 2(-9.8)(82.35)<br />
<br />
:v = - 40.2 m/s (negative because it is going down)<br />
<br />
:-(c) The total distance it traveled would be the distance it traveled going up, 7.35 m, and the total distance it went to the bottom of the cliff, 82.35 m. So adding these together, we get, 89.7 m.<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=115Giancoli Physics (5th ed) Chapter 22007-09-05T18:02:09Z<p>WikiSysop: /* 49. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
-The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at<br />
<br />
0 = 5.5 + (-9.8)t<br />
<br />
-5.5 = (-9.8)t<br />
<br />
t = .56 s<br />
<br />
-Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.<br />
<br />
02 = (5.5)2 +2(-9.8)s<br />
<br />
s = 1.54 m<br />
<br />
-Add this to the 105 m, and we get 106 .54 m.<br />
<br />
-How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2<br />
<br />
106.54 = (0)t + 1/2(9.8)t2<br />
<br />
t = 4.66 s<br />
<br />
-Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?===<br />
-We can use the formula s = ut + 1/2 at2, but the initial velocity (u) is zero, so we are left with, s = 1/2 at2. Replacing h with s, and solving for time gives us tf (the time to fall) = (2h)/g)^1/2<br />
<br />
-The time it takes for the sound to be heard up at the top of the cliff can be found by the formula v = s/t or t = s/v, which is ts = h /(340 m/s).<br />
<br />
-The total time is 3.4 s, so ts + tf = 3.4 s<br />
<br />
-Substitute for ts and tf. This gives us:<br />
<br />
((h /340) + (2h /g)^1/2 = 3.4)<br />
<br />
-This is the same as:<br />
<br />
(1/340) ((h)^1/2)2 + (2/g)^1/2 (h)^1/2 - 3.4 = 0<br />
<br />
-This is quadratic now, where x = h^1/2, so solve it using the quadratic formula (in your book on page 1042).<br />
<br />
h = 52 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?===<br />
-(a) If it is thrown upward, we must find the time it takes to stop going upward, and reach a final velocity (v) of zero. Using the formula v = u + at, find the time in the air.<br />
<br />
0 = 12 + (-9.8)t<br />
<br />
t = 1.23 s<br />
<br />
-Now find the height it went during that 1.22 s using the formula v2 = u2 + 2as.<br />
<br />
02 = (12)2 + 2(-9.8)s<br />
<br />
s = 7.35 m<br />
<br />
-Add 7.35 m to the height of the cliff, 75 m, to get 82.35 m. Now find the time it takes to fall this new height using the formula s = ut + 1/2 at2<br />
<br />
82.35 m = (0)t + 1/2(-9.8)t2<br />
<br />
t = 4.1 s<br />
<br />
-Add 4.1 s to the time the stone is going upward, which is 1.23 s, and you get 5.33 s.<br />
<br />
-(b) The initial velocity (u) is 0 m/s and the height is 82.35 m. Using the formula v2 = u2 + 2as, find the final velocity.<br />
<br />
v2 = (0)2 + 2(-9.8)(82.35)<br />
<br />
v = - 40.2 m/s (negative because it is going down)<br />
<br />
-(c) The total distance it traveled would be the distance it traveled going up, 7.35 m, and the total distance it went to the bottom of the cliff, 82.35 m. So adding these together, we get, 89.7 m.<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=114Giancoli Physics (5th ed) Chapter 22007-09-05T18:00:56Z<p>WikiSysop: /* 47. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
-The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at<br />
<br />
0 = 5.5 + (-9.8)t<br />
<br />
-5.5 = (-9.8)t<br />
<br />
t = .56 s<br />
<br />
-Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.<br />
<br />
02 = (5.5)2 +2(-9.8)s<br />
<br />
s = 1.54 m<br />
<br />
-Add this to the 105 m, and we get 106 .54 m.<br />
<br />
-How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2<br />
<br />
106.54 = (0)t + 1/2(9.8)t2<br />
<br />
t = 4.66 s<br />
<br />
-Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.A rock is dropped from a sea cliff and the sound of it striking the ocean is hears 3.4 seconds later. If the speed of sound is 340 m/s, how high is the cliff?===<br />
-We can use the formula s = ut + 1/2 at2, but the initial velocity (u) is zero, so we are left with, s = 1/2 at2. Replacing h with s, and solving for time gives us tf (the time to fall) = (2h)/g)^1/2<br />
<br />
-The time it takes for the sound to be heard up at the top of the cliff can be found by the formula v = s/t or t = s/v, which is ts = h /(340 m/s).<br />
<br />
-The total time is 3.4 s, so ts + tf = 3.4 s<br />
<br />
-Substitute for ts and tf. This gives us:<br />
<br />
((h /340) + (2h /g)^1/2 = 3.4)<br />
<br />
-This is the same as:<br />
<br />
(1/340) ((h)^1/2)2 + (2/g)^1/2 (h)^1/2 - 3.4 = 0<br />
<br />
-This is quadratic now, where x = h^1/2, so solve it using the quadratic formula (in your book on page 1042).<br />
<br />
h = 52 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=113Giancoli Physics (5th ed) Chapter 22007-09-05T17:58:51Z<p>WikiSysop: /* 41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the g</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
-The package and the helicopter are both 105 m above the earth, but are both going up at 5.5 m/s. First we must find out the amount of time the package spends going up, before it begins to go down to the earth. The final velocity (v) for when the package is going up is zero, because at this moment, the package isn't going up anymore, but is falling toward the ground. We can use the formula: v = u + at<br />
<br />
0 = 5.5 + (-9.8)t<br />
<br />
-5.5 = (-9.8)t<br />
<br />
t = .56 s<br />
<br />
-Now we know how long the package spends going up. But how high does the package go above the 105 m? Remember that the initial velocity (u) is zero. For this we can use the formula: v2 = u2 + 2as.<br />
<br />
02 = (5.5)2 +2(-9.8)s<br />
<br />
s = 1.54 m<br />
<br />
-Add this to the 105 m, and we get 106 .54 m.<br />
<br />
-How long does it take for the package to drop from this height, 106.54 m? We can use the formula: s = ut + 1/2 at2<br />
<br />
106.54 = (0)t + 1/2(9.8)t2<br />
<br />
t = 4.66 s<br />
<br />
-Remember that the package was going up first and took .56 s to do that. Add 4.66 s and .56 s to get 5.22 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=112Giancoli Physics (5th ed) Chapter 22007-09-05T17:58:08Z<p>WikiSysop: </p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=111Giancoli Physics (5th ed) Chapter 22007-09-05T17:57:01Z<p>WikiSysop: /* 40. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?===<br />
<br />
===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=110Giancoli Physics (5th ed) Chapter 22007-09-05T17:56:12Z<p>WikiSysop: /* 41. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41. A helicopter is ascending vertically with a speed of 5.5 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?<br />
===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=109Giancoli Physics (5th ed) Chapter 22007-09-05T17:55:40Z<p>WikiSysop: /* 37. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air before returning to Earth?===<br />
:-If the kangaroo jumps up, it must fall back to the ground. If it reaches a maximum height of 2.7 m, then we know that at this point, it is no longer going up, but is about to start falling back to the earth. At this point, at 2.7 m above the earth, we know that the final velocity (v) on the way up is zero, and the initial velocity (u) on the way down is also zero.<br />
<br />
:-We can use the formula: s = ut + 1/2 at2. Now put in the information we know.<br />
<br />
:2.7 = 0(t) + 1/2(9.8)t2<br />
<br />
:2.7 = 4.9t2<br />
<br />
:t = .74 s<br />
<br />
:-But since this is only on the way down, the time must be doubled to include the time it was going up.<br />
<br />
:.74 x 2 = 1.48 which is about 1.5 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=108Giancoli Physics (5th ed) Chapter 22007-09-05T17:53:52Z<p>WikiSysop: /* 35. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35. Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before "landing"?===<br />
:-We know that the initial velocity (u) of King Kong is 0 m/s, and that the height of the building is 380 m.<br />
<br />
:-(a) To find the time, we can use the formula: s = ut + 1/2 at2<br />
<br />
:380 m = 0(t) + 1/2 (9.8)t2<br />
<br />
:t2 = 77.6161<br />
<br />
:t = 8.81 s<br />
<br />
:-(b) To find the final velocity (v), we can use the formula: v2 = u2 + 2as<br />
<br />
:v2 = (0)2 + 2(9.8)(380)<br />
<br />
:86.3 m/s<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=107Giancoli Physics (5th ed) Chapter 22007-09-05T17:52:47Z<p>WikiSysop: /* 33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h? */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
:-First you have to change the 90 km/h to m/s<br />
<br />
::90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
:::= 25 m/s<br />
<br />
:-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
:-Using the formula v = u + at, we can find the time needed.<br />
<br />
::25 m/s = 0 + (9.8)t<br />
<br />
:t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=106Giancoli Physics (5th ed) Chapter 22007-09-05T17:52:07Z<p>WikiSysop: /* 33. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33. If a car rolls gently (v=0) off a vertical cliff, how long does it take to reach 90 km/h?===<br />
-First you have to change the 90 km/h to m/s<br />
<br />
90 km/h (1 hr / 3600 s)(1000 m / 1 km)<br />
<br />
= 25 m/s<br />
<br />
-You know that the initial velocity (u) is 0, and that the final velocity (v) is 25 m/s.<br />
<br />
-Using the formula v = u + at, we can find the time needed.<br />
<br />
25 m/s = 0 + (9.8)t<br />
<br />
t = 2.6 s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=105Giancoli Physics (5th ed) Chapter 22007-09-05T17:50:33Z<p>WikiSysop: /* 57. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57. Construct the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-26.===<br />
:-See the back of the text book for a graph. To construct a graph of velocity vs. time from a distance vs. time graph, you end up taking the derivative (if you have already learned derivatives) or you can take the slope of the distance at certain points and graph the different slopes, which ends up being the velocity.<br />
<br />
:The velocity graph is going to look like this:<br />
::interval Position Velocity<br />
::0-20 sec constant slope constant V (horizontal line)<br />
::20-30 sec curve concave upwards positive acceleration - upward sloping straight line<br />
::30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero<br />
::37-46 curve concave downwards negative acceleration, downward sloping line that continues down below zero (moving backwards)<br />
::46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=104Giancoli Physics (5th ed) Chapter 22007-09-05T17:48:37Z<p>WikiSysop: /* 53. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53. Figure 2-27 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?===<br />
:-(a) Looking at the graph of v vs. time, we can see that the greatest velocity was achieved around 50 s.<br />
<br />
:-(b) The velocity is constant when the velocity doesn't change, for if it does change, then there is an acceleration. The time period of 90-107 seconds has a constant velocity. (it is also 0)<br />
<br />
:-(c) the acceleration is constant when the velocity either increases or decreases at a constant rate. These look like straight lines on a Velocity vs time graph. From 0 to 20 s and from 90 to 107 s (the acceleration was constant, 0)<br />
<br />
:-(d) The magnitude of acceleration was greatest when the slope of the velocity was steepest. This occurs at about 75 s.<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=103Giancoli Physics (5th ed) Chapter 22007-09-05T17:46:54Z<p>WikiSysop: /* 51. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51. The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-26. What is its instantaneous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?===<br />
:-(a) Make a tangent line at t = 10 s to find the instantaneous velocity by its slope. This is the less steep of the two blue lines above. This line seems to have a slope of rise/run = 14 m/50 s = .28 m/s<br />
<br />
:-(b) Make a tangent line at t = 30 s, and find its slope (this is the velocity). This is the steeper of the two blue lines. This line rises 25 m between 17 and 37 seconds, so it has a slope of about 25 m/20 s, which is about 1.25 or 1.2 m/s.<br />
<br />
:-(c) Average velocity is just total displacement divided by total time. From 0 to 5 seconds it seems to go from 0 to about 1.5 m (read from the graph), and the time is of course 5 seconds, so the average velocity is 1.5 m/5 = .3 m/s<br />
<br />
:-(d) From 25 to 30 the rabbit displaces from 8 m to 16 m (read from the graph), again in 5 seconds giving an average velocity of 8m/5 s = 1.6 m/s<br />
<br />
:-(e) From 40 to 50 seconds the rabbit displaces from 20 m to 10 m (read from the graph), now in 10 seconds giving an average velocity of -10m/10s = -1.0 m/s<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=102Giancoli Physics (5th ed) Chapter 22007-09-05T17:45:13Z<p>WikiSysop: /* 23. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23. A car slows down from a speed of 25.0 m/s to rest in 5.00 s. How far did it travel in this time?===<br />
:-To start, it is helpful to figure out all that the question just gave us. We know that the initial velocity is 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its not moving). It takes a total of 5 seconds to do this. Also note that it is slowing down, so the acceleration is negative.<br />
<br />
:-We must first find what the acceleration is by using this formula: aav = Dv/Dt , then put the numbers in and solve.<br />
<br />
::a= (25 m/s) / (5 sec)<br />
::a= - 5 m/s/s, because it is slowing down<br />
<br />
:-Now that we know what the acceleration is, we can solve for the distance it takes to go from 25 m/s to 0 m/s in 6 seconds. For this we can use the formula: v2 = u2 + 2as, then put the numbers in and solve.<br />
<br />
::(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s<br />
<br />
::0= 625 + (-10) s<br />
<br />
::-625= (-10) s<br />
<br />
:s= 62.5 m<br />
<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=101Giancoli Physics (5th ed) Chapter 22007-09-05T17:43:16Z<p>WikiSysop: /* 21. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21. A light plane must reach a speed of 30 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s/s?===<br />
:-The plane is starting from rest, and therefore, its initial velocity, u, is 0 m/s. This helps out a great deal to note this. The final velocity, v, is 30 m/s. The acceleration is 3.0 m/s/s.<br />
<br />
:-A formula we can use is v2= u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Now put the numbers in:<br />
::(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s<br />
<br />
:The initial velocity here is zero, so we can drop it from the equation.<br />
<br />
::900= 2(3.0 m/s/s) s<br />
<br />
:s= 1.5x102 m<br />
<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Blank_Problem_Template&diff=100Blank Problem Template2007-09-05T17:34:26Z<p>WikiSysop: </p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=99Giancoli Physics (5th ed) Chapter 22007-09-05T17:33:14Z<p>WikiSysop: /* 16. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:(-6.3ms<sup>-2</sup>, .64 g)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=98Giancoli Physics (5th ed) Chapter 22007-09-05T17:32:23Z<p>WikiSysop: /* 14. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:(5.2 s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=97Giancoli Physics (5th ed) Chapter 22007-09-05T17:31:47Z<p>WikiSysop: /* 12. */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:(6.73 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=96Giancoli Physics (5th ed) Chapter 22007-09-05T17:31:22Z<p>WikiSysop: /* Problems */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:(105 km/h, 29 m/s, 95 ft/s)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:(about 300 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:(10.4 m/s, +3.5 m/s)<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:(4.43 h, 881 km/h)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=95Giancoli Physics (5th ed) Chapter 22007-09-05T17:29:26Z<p>WikiSysop: /* Problems */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/hr)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=94Giancoli Physics (5th ed) Chapter 22007-09-05T17:27:31Z<p>WikiSysop: /* Problems */</p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
:(70.8 km/hr)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
:()<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=93Giancoli Physics (5th ed) Chapter 22007-09-05T17:26:30Z<p>WikiSysop: </p>
<hr />
<div>[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
==Problems==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=92Giancoli Physics (5th ed) Chapter 22007-09-05T17:26:04Z<p>WikiSysop: </p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=91Giancoli Physics (5th ed) Chapter 22007-09-05T17:25:07Z<p>WikiSysop: /* Problems */</p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
== Problems ==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=90Giancoli Physics (5th ed) Chapter 22007-09-05T17:23:35Z<p>WikiSysop: </p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
== Problems ==<br />
<br />
===1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?===<br />
:Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:<br />
:F = ma<br />
:F = (60.0 kg)(1.15 m/s/s) = 69 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?=== <br />
:They give us m = 9.0 g = 9.0 x 10<sup>-3</sup> kg, and a = 10,000 "g's" = 98000 m/s/s so:<br />
:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===4. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?=== <br />
:Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:<br />
:on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N<br />
:on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N<br />
:on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N<br />
:At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N<br />
:(Since the velocity is constant, there is no acceleration so they are weightless)<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===6. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?===<br />
:First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:<br />
::90 km/hr/3.6 = 25 m/s<br />
:We have a cute linear kinematics problem to solve:<br />
:s = don't care<br />
:u = 90 km/hr/3.6 = 25 m/s<br />
:v = 0<br />
:a = ???<br />
:t = 8.0 s <br />
<br />
:Use v = u + at to find a:<br />
:v = u + at<br />
:0 = 25 m/s + a(8.0 s)<br />
:a = -3.125 m/s/s<br />
:so now we can use Newton's second law<br />
:F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N<br />
[[#top | Table of Contents]]<br />
----<br />
===8. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s<sup>2</sup> using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish? ===<br />
:The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:<br />
:<22 N - mg> = m(+4.5 m/s/s)<br />
:<22 - 9.8m> = m(4.5)<br />
:22 = 9.8m + 4.5m = 14.3m<br />
:m = 22/14.3 = 1.538 kg = 1.5 kg<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===10. ===<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s? ===<br />
<br />
:We have a cute linear kinematics problem to solve first:<br />
:s = 2.8 m<br />
:u = 0 (assumed)<br />
:v = 13 m/s<br />
:a = ???<br />
:t = Don't care <br />
<br />
:Use v2 = u2 + 2as, a = 30.1786 m/s/s<br />
:F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N<br />
[[#top | Table of Contents]]<br />
----<br />
===12. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down? ===<br />
<br />
:First calculate the weight:<br />
:F = ma = (10 kg)(9.80 m/s/s) = 98 N <br />
<br />
:Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:<br />
:<63 N - 98 N> = (10 kg) a<br />
:a = -3.5 m/s/s (down)<br />
[[#top | Table of Contents]]<br />
----<br />
===14. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.=== <br />
<br />
:If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.<br />
:Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:<br />
:<568.4 N - 735 N> = (75 kg)a<br />
:a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===16. ===<br />
[[#top | Table of Contents]]<br />
----<br />
===17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking? ===<br />
:First calculate the weight:<br />
:F = ma = (2100 kg)(9.80 m/s/s) = 20580 N<br />
<br />
:Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:<br />
:<21750 N - 20580 N> = (2100 kg) a<br />
:a = +.557 m/s/s = +.56 m/s/s upward<br />
[[#top | Table of Contents]]<br />
----<br />
<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Solutions&diff=89Giancoli Physics (5th ed) Solutions2007-09-05T17:22:10Z<p>WikiSysop: </p>
<hr />
<div>[[Main Page | Return to Main Page]]<br />
<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 1]]<br />
*[[Giancoli Physics (5th ed) Chapter 2 | Chapter 2]]<br />
*[[Giancoli Physics (5th ed) Chapter 3 | Chapter 3]]<br />
*[[Giancoli Physics (5th ed) Chapter 4 | Chapter 4]]<br />
*[[Giancoli Physics (5th ed) Chapter 5 | Chapter 5]]<br />
<br />
<br />
[[Blank Problem Template]] - Has all the links and sections to make a solutions page for the chapter</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Solutions&diff=88Giancoli Physics (5th ed) Solutions2007-09-05T17:21:30Z<p>WikiSysop: </p>
<hr />
<div>[[Main Page | Return to Main Page]]<br />
<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 1]]<br />
*[[Giancoli Physics (5th ed) Chapter 2 | Chapter 2]]<br />
*[[Giancoli Physics (5th ed) Chapter 3 | Chapter 3]]<br />
*[[Giancoli Physics (5th ed) Chapter 4 | Chapter 4]]<br />
*[[Giancoli Physics (5th ed) Chapter 5 | Chapter 5]]<br />
<br />
<br />
[[Blank Solution Template]] - Has all the links and sections to make a solutions page for the chapter</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Solutions&diff=87Giancoli Physics (5th ed) Solutions2007-09-05T17:19:52Z<p>WikiSysop: </p>
<hr />
<div>[[Main Page | Return to Main Page]]<br />
<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 1]]<br />
*[[Giancoli Physics (5th ed) Chapter 2 | Chapter 2]]<br />
*[[Giancoli Physics (5th ed) Chapter 3 | Chapter 3]]<br />
*[[Giancoli Physics (5th ed) Chapter 4 | Chapter 4]]<br />
*[[Giancoli Physics (5th ed) Chapter 5 | Chapter 5]]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_5&diff=86Giancoli Physics (5th ed) Chapter 52007-09-05T17:18:18Z<p>WikiSysop: New page: ==Problems== Return to Solutions ===1.=== Table of Contents ---- ===2.=== Table of Contents ---- ===3.=== [[#top | Table of ...</p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_4&diff=85Giancoli Physics (5th ed) Chapter 42007-09-05T17:18:07Z<p>WikiSysop: New page: ==Problems== Return to Solutions ===1.=== Table of Contents ---- ===2.=== Table of Contents ---- ===3.=== [[#top | Table of ...</p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_3&diff=84Giancoli Physics (5th ed) Chapter 32007-09-05T17:17:56Z<p>WikiSysop: New page: ==Problems== Return to Solutions ===1.=== Table of Contents ---- ===2.=== Table of Contents ---- ===3.=== [[#top | Table of ...</p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_2&diff=83Giancoli Physics (5th ed) Chapter 22007-09-05T17:17:35Z<p>WikiSysop: New page: ==Problems== Return to Solutions ===1.=== Table of Contents ---- ===2.=== Table of Contents ---- ===3.=== [[#top | Table of ...</p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Solutions&diff=82Giancoli Physics (5th ed) Solutions2007-09-05T17:17:16Z<p>WikiSysop: </p>
<hr />
<div>*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 1]]<br />
*[[Giancoli Physics (5th ed) Chapter 2 | Chapter 2]]<br />
*[[Giancoli Physics (5th ed) Chapter 3 | Chapter 3]]<br />
*[[Giancoli Physics (5th ed) Chapter 4 | Chapter 4]]<br />
*[[Giancoli Physics (5th ed) Chapter 5 | Chapter 5]]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_1&diff=81Giancoli Physics (5th ed) Chapter 12007-09-05T17:16:03Z<p>WikiSysop: </p>
<hr />
<div>==Problems==<br />
[[Giancoli Physics (5th ed) Solutions | Return to Solutions]]<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Chapter_1&diff=80Giancoli Physics (5th ed) Chapter 12007-09-05T17:14:48Z<p>WikiSysop: New page: ==Problems== ===1.=== Table of Contents ---- ===2.=== Table of Contents ---- ===3.=== Table of Contents ---- ===4.=== Table of Contents ---- ===...</p>
<hr />
<div>==Problems==<br />
===1.===<br />
[[#top | Table of Contents]]<br />
----<br />
===2.===<br />
[[#top | Table of Contents]]<br />
----<br />
===3.===<br />
[[#top | Table of Contents]]<br />
----<br />
===4.===<br />
[[#top | Table of Contents]]<br />
----<br />
===5.===<br />
[[#top | Table of Contents]]<br />
----<br />
===6.===<br />
[[#top | Table of Contents]]<br />
----<br />
===7.===<br />
[[#top | Table of Contents]]<br />
----<br />
===8.===<br />
[[#top | Table of Contents]]<br />
----<br />
===9.===<br />
[[#top | Table of Contents]]<br />
----<br />
===10.===<br />
[[#top | Table of Contents]]<br />
----<br />
===11.===<br />
[[#top | Table of Contents]]<br />
----<br />
===12.===<br />
[[#top | Table of Contents]]<br />
----<br />
===13.===<br />
[[#top | Table of Contents]]<br />
----<br />
===14.===<br />
[[#top | Table of Contents]]<br />
----<br />
===15.===<br />
[[#top | Table of Contents]]<br />
----<br />
===16.===<br />
[[#top | Table of Contents]]<br />
----<br />
===17.===<br />
[[#top | Table of Contents]]<br />
----<br />
===18.===<br />
[[#top | Table of Contents]]<br />
----<br />
===19.===<br />
[[#top | Table of Contents]]<br />
----<br />
===20.===<br />
[[#top | Table of Contents]]<br />
----<br />
===21.===<br />
[[#top | Table of Contents]]<br />
----<br />
===22.===<br />
[[#top | Table of Contents]]<br />
----<br />
===23.===<br />
[[#top | Table of Contents]]<br />
----<br />
===24.===<br />
[[#top | Table of Contents]]<br />
----<br />
===25.===<br />
[[#top | Table of Contents]]<br />
----<br />
===26.===<br />
[[#top | Table of Contents]]<br />
----<br />
===27.===<br />
[[#top | Table of Contents]]<br />
----<br />
===28.===<br />
[[#top | Table of Contents]]<br />
----<br />
===29.===<br />
[[#top | Table of Contents]]<br />
----<br />
===30.===<br />
[[#top | Table of Contents]]<br />
----<br />
===31.===<br />
[[#top | Table of Contents]]<br />
----<br />
===32.===<br />
[[#top | Table of Contents]]<br />
----<br />
===33.===<br />
[[#top | Table of Contents]]<br />
----<br />
===34.===<br />
[[#top | Table of Contents]]<br />
----<br />
===35.===<br />
[[#top | Table of Contents]]<br />
----<br />
===36.===<br />
[[#top | Table of Contents]]<br />
----<br />
===37.===<br />
[[#top | Table of Contents]]<br />
----<br />
===38.===<br />
[[#top | Table of Contents]]<br />
----<br />
===39.===<br />
[[#top | Table of Contents]]<br />
----<br />
===40.===<br />
[[#top | Table of Contents]]<br />
----<br />
===41.===<br />
[[#top | Table of Contents]]<br />
----<br />
===42.===<br />
[[#top | Table of Contents]]<br />
----<br />
===43.===<br />
[[#top | Table of Contents]]<br />
----<br />
===44.===<br />
[[#top | Table of Contents]]<br />
----<br />
===45.===<br />
[[#top | Table of Contents]]<br />
----<br />
===46.===<br />
[[#top | Table of Contents]]<br />
----<br />
===47.===<br />
[[#top | Table of Contents]]<br />
----<br />
===48.===<br />
[[#top | Table of Contents]]<br />
----<br />
===49.===<br />
[[#top | Table of Contents]]<br />
----<br />
===50.===<br />
[[#top | Table of Contents]]<br />
----<br />
===51.===<br />
[[#top | Table of Contents]]<br />
----<br />
===52.===<br />
[[#top | Table of Contents]]<br />
----<br />
===53.===<br />
[[#top | Table of Contents]]<br />
----<br />
===54.===<br />
[[#top | Table of Contents]]<br />
----<br />
===55.===<br />
[[#top | Table of Contents]]<br />
----<br />
===56.===<br />
[[#top | Table of Contents]]<br />
----<br />
===57.===<br />
[[#top | Table of Contents]]<br />
----<br />
===58.===<br />
[[#top | Table of Contents]]<br />
----<br />
===59.===<br />
[[#top | Table of Contents]]<br />
----<br />
===60.===<br />
[[#top | Table of Contents]]<br />
----<br />
===61.===<br />
[[#top | Table of Contents]]<br />
----<br />
===62.===<br />
[[#top | Table of Contents]]<br />
----<br />
===63.===<br />
[[#top | Table of Contents]]<br />
----<br />
===64.===<br />
[[#top | Table of Contents]]<br />
----<br />
===65.===<br />
[[#top | Table of Contents]]<br />
----<br />
===66.===<br />
[[#top | Table of Contents]]<br />
----<br />
===67.===<br />
[[#top | Table of Contents]]<br />
----<br />
===68.===<br />
[[#top | Table of Contents]]<br />
----<br />
===69.===<br />
[[#top | Table of Contents]]<br />
----<br />
===70.===<br />
[[#top | Table of Contents]]<br />
----<br />
===71.===<br />
[[#top | Table of Contents]]<br />
----<br />
===72.===<br />
[[#top | Table of Contents]]<br />
----<br />
===73.===<br />
[[#top | Table of Contents]]<br />
----<br />
===74.===<br />
[[#top | Table of Contents]]<br />
----<br />
===75.===<br />
[[#top | Table of Contents]]<br />
----<br />
===76.===<br />
[[#top | Table of Contents]]<br />
----<br />
===77.===<br />
[[#top | Table of Contents]]<br />
----<br />
===78.===<br />
[[#top | Table of Contents]]<br />
----<br />
===79.===<br />
[[#top | Table of Contents]]<br />
----<br />
===80.===<br />
[[#top | Table of Contents]]<br />
----<br />
===81.===<br />
[[#top | Table of Contents]]<br />
----<br />
===82.===<br />
[[#top | Table of Contents]]<br />
----<br />
===83.===<br />
[[#top | Table of Contents]]<br />
----<br />
===84.===<br />
[[#top | Table of Contents]]<br />
----<br />
===85.===<br />
[[#top | Table of Contents]]<br />
----<br />
===86.===<br />
[[#top | Table of Contents]]<br />
----<br />
===87.===<br />
[[#top | Table of Contents]]<br />
----<br />
===88.===<br />
[[#top | Table of Contents]]<br />
----<br />
===89.===<br />
[[#top | Table of Contents]]<br />
----<br />
===90.===<br />
[[#top | Table of Contents]]<br />
----<br />
===91.===<br />
[[#top | Table of Contents]]<br />
----<br />
===92.===<br />
[[#top | Table of Contents]]<br />
----<br />
===93.===<br />
[[#top | Table of Contents]]<br />
----<br />
===94.===<br />
[[#top | Table of Contents]]<br />
----<br />
===95.===<br />
[[#top | Table of Contents]]<br />
----<br />
===96.===<br />
[[#top | Table of Contents]]<br />
----<br />
===97.===<br />
[[#top | Table of Contents]]<br />
----<br />
===98.===<br />
[[#top | Table of Contents]]<br />
----<br />
===99.===<br />
[[#top | Table of Contents]]<br />
----<br />
===100.===<br />
[[#top | Table of Contents]]<br />
----</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Giancoli_Physics_(5th_ed)_Solutions&diff=79Giancoli Physics (5th ed) Solutions2007-09-05T17:13:05Z<p>WikiSysop: New page: * Chapter 1 * Chapter 2 * Chapter 3 *[[Giancoli Physics (5th ed) ...</p>
<hr />
<div>*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 1]]<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 2]]<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 3]]<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 4]]<br />
*[[Giancoli Physics (5th ed) Chapter 1 | Chapter 5]]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Main_Page&diff=78Main Page2007-09-05T17:09:33Z<p>WikiSysop: </p>
<hr />
<div><big>Welcome to the Tualatin High School Physics Wiki</big><br />
<br />
<br />
<br />
IB Physics - Douglas Giancoli's Physics - 5th edition<br />
:[[Giancoli Physics (5th ed) Solutions | Solutions to Problems]]</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=Main_Page&diff=77Main Page2007-09-05T17:05:28Z<p>WikiSysop: </p>
<hr />
<div><big>Welcome to the Tualatin High School Physics Wiki</big><br />
<br />
<br />
<br />
IB Physics - Douglas Giancoli's Physics - 5th edition</div>WikiSysophttps://tuhsphysics.ttsd.k12.or.us/wiki/index.php?title=User:Cmurray&diff=76User:Cmurray2007-09-05T04:27:27Z<p>WikiSysop: </p>
<hr />
<div>Here is the original first page:<br />
<br />
<big>'''MediaWiki has been successfully installed.'''</big><br />
<br />
Consult the [http://meta.wikimedia.org/wiki/Help:Contents User's Guide] for information on using the wiki software.<br />
<br />
== Getting started ==<br />
<br />
* [http://www.mediawiki.org/wiki/Help:Configuration_settings Configuration settings list]<br />
* [http://www.mediawiki.org/wiki/Help:FAQ MediaWiki FAQ]<br />
* [http://mail.wikimedia.org/mailman/listinfo/mediawiki-announce MediaWiki release mailing list]<br />
<br />
[[Murray's Sand Box]]</div>WikiSysop