Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 14 - Exercises and Problems: 34

Answer

Please see the work below.

Work Step by Step

We know that $v=\sqrt{\frac{\gamma P}{\rho}}$ This simplifies to: $\gamma=\sqrt{\frac{v^2 \rho}{P}}$ We plug in the known values to obtain: $\gamma=\frac{(368\frac{m}{s})^2\times 1\frac{Kg}{m^3}}{81000Pa}=1.67=\frac{5}{3}$
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