University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises: 1.30

Answer

(a) $A_y =−23.7~ m$ (b) $A=28.6 ~m$

Work Step by Step

(a) We can find the magnitude of $A_y$. $\frac{A_x}{A_y}= tan(\theta)$ $A_y= \frac{A_x}{tan(\theta)} = \frac{16.0~m}{tan(34^{\circ})} = 23.7 ~m$ Since the angle is clockwise from the negative y-axis, then $A_y =−23.7~ m$ (b) We can use $A_x$ and $A_y$ to find the magnitude of $A$. $A^2=(A_x)^2+(A_y)^2$ $A=\sqrt{(−16.0 ~m)^2+(-23.7 ~m)^2}$ $A=28.6 ~m$
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