University Physics with Modern Physics (14th Edition)

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

Chapter 37 - Relativity - Problems - Exercises - Page 1249: 37.30


See explanation.

Work Step by Step

a. This is nonrelativistic. F = ma. $$a=\frac{F}{m}=5.49\times10^{15}m/s^2$$ b. According to equation 37.32, in this situation, the force is $F=\gamma^3 ma$. The value of $\gamma$ is $\frac{1}{\sqrt{1-v^{2}/c^{2}}}=1.81$. $$a=\frac{F}{m \gamma^3}=9.27\times10^{14}m/s^2$$
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