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 1145: 2c

Answer

$\beta = 0.99995000 $

Work Step by Step

lorentz factor $\gamma=100$ We know that the formula for lorentz factor is; $\gamma =\frac{1}{\sqrt (1-\frac{1}{\beta^2})}$ The speed parameter $\beta =\sqrt (1-\frac{1}{\gamma^2})$ We substitute $\gamma$ value i,e 100 in above formula and solve; $\beta =\sqrt (1-\frac{1}{100^2})$ $\beta = 0.99995000 $
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