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 36 - Relativity - Exercises and Problems - Page 1059: 21

Answer

$v = 0.80 c$

Work Step by Step

Let $L = 0.60~L_0$. We can find the speed: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{L}{L_0}$ $1-\frac{v^2}{c^2} = (\frac{L}{L_0})^2$ $\frac{v^2}{c^2} = 1-(\frac{L}{L_0})^2$ $v^2 = [1-(\frac{L}{L_0})^2]~\times c^2$ $v = \sqrt{1-(\frac{L}{L_0})^2}~\times c$ $v = \sqrt{1-(\frac{0.60~L_0}{L_0})^2}~\times c$ $v = \sqrt{1-(0.60)^2}~\times c$ $v = 0.80 c$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.