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 1060: 48

Answer

The tube has a length of $~~78.4~cm~~$ in the electrons' reference frame.

Work Step by Step

We can find $\gamma$: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{(0.99999997~c)^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{0.0000000599999991}}$ $\gamma = 4082.5$ Let $L_0 = 3.2~km$ We can find the length $L$ of the tube in the electrons' reference frame: $L = \frac{L_0}{\gamma}$ $L = \frac{3.2~km}{4082.5}$ $L =0.000784~km$ $L = 78.4~cm$ The tube has a length of $~~78.4~cm~~$ in the electrons' reference frame.
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