Difference between revisions of "Skill Set 03.1"

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(2. Find the components of the vector.)
(3. Question)
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===3. Question===
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===3. Convert 13.2 m/s x + 5.70 m/s y into an angle magnitude vector.===
 
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<blockquote>
Solution goes here
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To get the angle:<br><br>
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tan<sup>-1</sup>(5.7/13.2) = 23.4<sup>o</sup><br><br>
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To get the magnitude:<br><br>
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a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup><br>
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5.7<sup>2</sup> + 13.2 <sup>2</sup> = c<sup>2</sup><br>
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c = 14.4 m<br><br>
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'''14.4 m @ 23.4<sup>o</sup>'''
 
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[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
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===4. Question===
 
===4. Question===
 
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Revision as of 10:10, 12 February 2009

Main Page > IB Physics Skill Sets > Skill Set 03.1

1. Find the components of the vector.

Since the angle given is the same as the trig angle, we can use 17o in our equations.

32cos(17) = 30.60 m x
32sin(17) = 9.356 m y

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2. Find the components of the vector.

Converting the 23o angle to a trig angle is done by subtracting it by 360, since the angle is just below the 0 or 360o line. Using our new angle, 337o, we can solve for the components.

15cos(337) = 13.81 m x
15sin(337) = -5.861 m y

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3. Convert 13.2 m/s x + 5.70 m/s y into an angle magnitude vector.

To get the angle:

tan-1(5.7/13.2) = 23.4o

To get the magnitude:

a2 + b2 = c2
5.72 + 13.2 2 = c2
c = 14.4 m

14.4 m @ 23.4o

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4. Question

Solution goes here

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5. Question

Solution goes here

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