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 4 - Kinematics in Two Dimensions - Exercises and Problems: 28

Answer

(a) $v = 418~m/s$ (b) $v = 197~m/s$

Work Step by Step

We can find the angular speed of the earth as it rotates once each day. $\omega = \frac{2\pi~rad}{(24)(3600~s)}$ $\omega = 7.27\times 10^{-5}~rad/s$ Note that for a point on the earth's surface at a latitude of $\theta$, the radius of rotation is $(6.4\times 10^6~m)~cos(\theta)$ (a) $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.4\times 10^6~m)~cos(26^{\circ})$ $v = 418~m/s$ (b) $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.4\times 10^6~m)~cos(65^{\circ})$ $v = 197~m/s$
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