Difference between revisions of "Skill Set 03.1"
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− | ===3. | + | ===3. Convert 13.2 m/s x + 5.70 m/s y into an angle magnitude vector.=== |
<blockquote> | <blockquote> | ||
− | + | To get the angle:<br><br> | |
+ | |||
+ | tan<sup>-1</sup>(5.7/13.2) = 23.4<sup>o</sup><br><br> | ||
+ | |||
+ | To get the magnitude:<br><br> | ||
+ | |||
+ | a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup><br> | ||
+ | 5.7<sup>2</sup> + 13.2 <sup>2</sup> = c<sup>2</sup><br> | ||
+ | c = 14.4 m<br><br> | ||
+ | |||
+ | '''14.4 m @ 23.4<sup>o</sup>''' | ||
</blockquote> | </blockquote> | ||
[[#top | Table of Contents]] | [[#top | Table of Contents]] | ||
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===4. Question=== | ===4. Question=== | ||
<blockquote> | <blockquote> |
Revision as of 10:10, 12 February 2009
Main Page > IB Physics Skill Sets > Skill Set 03.1
Contents
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
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
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
4. Question
Solution goes here
5. Question
Solution goes here