Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 15 - Exercises and Problems - Page 289: 16

Answer

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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