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 291: 64

Answer

$F=503MN$

Work Step by Step

We know that $P=P_{\circ}+\rho gh$ We plug in the known values to obtain: $P=101325+10^3\times 9.8\times 11\times 10^3=108MPa$ We also know that $R_{\circ}=\frac{D_{\circ}}{2}=\frac{121.8}{2}=60.9cm$ As $P=\frac{F}{A}$ $\implies F=PA$ We plug in the known values to obtain: $F=108MPa\times 4\pi R_{\circ}^2$ $F=108\times 10^6\times 4(3.1416)\times (0.609)^2$ $F=503MN$
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