Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 15 - Exercises and Problems - Page 281: 70


The answer is below.

Work Step by Step

a) We are told to integrate the following equation, so we find: $p = \rho h_0g\\ \int dP = \int -P\frac{dh}{h_0}$ We know that the bounds of the left hand integral are from $p_0$ to p, and the bounds of the right integral are from $h_0$ to h. Thus, we find: $ln(\frac{p}{p_0})=\frac{-h}{h_0}\\ p = p_0e^{\frac{-h}{h_0}}$ b) We find: $.5p_0=p_0e^{\frac{-h}{h_0}}$ $ h = -h_0ln(.5) = -(8.2)(ln.5) = 5.68 \ km$
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