Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 37 - Relativity - Problems - Page 1148: 54b

Answer

$\beta = 0.866$

Work Step by Step

We can find $\gamma$: $E = \gamma~mc^2 = 2.00~E_0$ $\gamma~E_0 = 2.00~E_0$ $\gamma = 2.00$ We can find $\beta$: $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ $1-\beta^2 = \frac{1}{\gamma^2}$ $\beta^2 = 1-\frac{1}{\gamma^2}$ $\beta = \sqrt{1-\frac{1}{\gamma^2}}$ $\beta = \sqrt{1-\frac{1}{(2.00)^2}}$ $\beta = 0.866$
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.