Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 29 - Relativity - Problems and Conceptual Exercises - Page 1045: 94

Answer

$199~pulses/s$

Work Step by Step

We know that $f^{\prime}=\frac{f}{\sqrt{1-\frac{v^2}{c^2}}}\sqrt{1-\frac{1}{c^2}[\frac{v-v_1}{1-\frac{vv_1}{c^2}}]^2}$ We plug in the known values to obtain: $f^{\prime}=\frac{153}{\sqrt{1-(\frac{0.800c}{c})^2}}\sqrt{1-\frac{1}{c^2}[\frac{0.800c-0.950c}{1-\frac{(0.800c)(0.950c)}{c^2}}]^2}$ This simplifies to: $f^{\prime}=199~pulses/s$
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