Difference between revisions of "Skill Set 03.1"

From TuHSPhysicsWiki
Jump to: navigation, search
(2. Question)
(2. Find the components of the vector.)
Line 16: Line 16:
 
Converting the 23<sup>o</sup> angle to a trig angle is done by subtracting it by 360, since the angle is just below the 0 or 360<sup>o</sup> line. Using our new angle, 337<sup>o</sup>, we can solve for the components.<br><br>
 
Converting the 23<sup>o</sup> angle to a trig angle is done by subtracting it by 360, since the angle is just below the 0 or 360<sup>o</sup> line. Using our new angle, 337<sup>o</sup>, we can solve for the components.<br><br>
  
15cos(337) = '''13.81 m x'''
+
15cos(337) = '''13.81 m x'''<br>
 
15sin(337) = '''-5.861 m y'''
 
15sin(337) = '''-5.861 m y'''
 
</blockquote>
 
</blockquote>

Revision as of 10:07, 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

Table of Contents


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

Table of Contents


3. Question

Solution goes here

Table of Contents


4. Question

Solution goes here

Table of Contents


5. Question

Solution goes here

Table of Contents