## University Physics with Modern Physics (14th Edition)

(a) The gauge pressure at a depth of 250 meters is $2.52\times 10^6~N/m^2$. (b) The net force on the window is $1.78\times 10^5~N$.
(a) We can find the gauge pressure at a depth of 250 meters. $P_G = \rho~g~h$ $P_G = (1030~kg/m^3)(9.80~m/s^2)(250~m)$ $P_G = 2.52\times 10^6~N/m^2$ The gauge pressure at a depth of 250 meters is $2.52\times 10^6~N/m^2$. (b) The difference in pressure on the outside and inside of the window is the gauge pressure. We can find the net force on the window. $F = P_G~A$ $F = (2.52\times 10^6~N/m^2)(\pi)(0.150~m)^2$ $F = 1.78\times 10^5~N$ The net force on the window is $1.78\times 10^5~N$.