Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems: 8

Answer

The magnitude of C is 3.15 m and the magnitude of D is 25.6 m. $C_x = -3.04~m$ $C_y = 0.815~m$ $D_x = 12.8~m$ $D_y = -22.2~m$

Work Step by Step

The magnitude of C is 3.15 m and the magnitude of D is 25.6 m. Therefore; $C_x = -C~cos(\theta)$ $C_x = -(3.15~m)~cos(15^{\circ})$ $C_x = -3.04~m$ $C_y = C~sin(\theta)$ $C_y = (3.15~m)~sin(15^{\circ})$ $C_y = 0.815~m$ $D_x = D~sin(\theta)$ $D_x = (25.6~m)~sin(30^{\circ})$ $D_x = 12.8~m$ $D_y = -D~cos(\theta)$ $D_y = -(25.6~m)~cos(30^{\circ})$ $D_y = -22.2~m$
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