Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 437: 35

Answer

$27psi$

Work Step by Step

We can find the required pressure in tire as follows: $F=\frac{mg}{4}$ $\implies F=\frac{1500(9.8)}{4}=3675N$ $A=0.13\times 0.15=0.0195m^2$ Now $P=\frac{F}{A}$ We plug in the known values to obtain: $P=\frac{3675}{0.0195}$ $\implies P=188KPa=27psi$
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