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 279: 22


Please see the work below.

Work Step by Step

We know that the area is $A=\pi r^2$ $A=(3.1416)(\frac{0.0015}{2})^2=7.67\times 10^{-6}m^2$ The pressure is $P=120atm=(120)(1.01\times 10^5)Pa=1.212\times 10^7Pa$ Now we can find the force as $F=PA$ $F=(1.212\times 10^7)(1.767\times 10^{-6})$ $F=21N$
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