Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - Problems - Page 767: 2

Answer

$2.07\times10^{-6}s$

Work Step by Step

You measure the dilated time interval. Find the proper time interval using equation 26–1a. $$\Delta t_o = \Delta t \sqrt{1-\frac{ v^{2} } { c^{2} }} $$ $$= (4.76\times10^{-6}s) \sqrt{1-\frac{ (2.70\times10^8m/s)^{2} } { (3.00\times10^8m/s)^{2} }} $$ $$= 2.07\times10^{-6}s$$
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