Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 14 - Fluids and Elasticity - Exercises and Problems: 10

Answer

The submarine's maximum safe depth in seawater is 3.16 km

Work Step by Step

We can find the area of the window as: $A = \pi~r^2$ $A = \pi~(0.10~m)^2$ $A = 0.0314~m^2$ We can find the pressure difference the window can withstand as: $\Delta P = \frac{F}{A}$ $\Delta P = \frac{1.0\times 10^6~N}{0.0314~m^2}$ $\Delta P = 3.185\times 10^7~N/m^2$ We can find the maximum safe pressure $P$ outside the window if the pressure inside $P_i$ is 1 atm. $\Delta P = P-P_i$ $P = \Delta P+P_i$ $P = 3.185\times 10^7~N/m^2+1.013\times 10^5~N/m^2$ $P = 3.195\times 10^7~N/m^2$ We can find the depth $h$ below sea level which has this pressure. $P = P_0 + \rho~g~h$ $h = \frac{P-P_0}{\rho~g}$ $h = \frac{(3.195\times 10^7~N/m^2)-(1.013\times 10^5~N/m^2)}{(1030~kg/m^3)(9.80~m/s^2)}$ $h = 3160~m = 3.16~km$ The submarine's maximum safe depth in seawater is 3.16 km.
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