## College Physics (4th Edition)

(a) The water pressure at a depth of 10.915 km is $1.10\times 10^8~Pa$ (b) The force exerted on the top of the hull due to the water pressure is $1.10\times 10^8~N$
(a) We can find the water pressure at a depth of 10.915 km: $P_{abs} = P_{atm}+P_g$ $P_{abs} = P_{atm}+\rho~g~h$ $P_{abs} = 101~kPa+(1025~kg/m^3)(9.80~m/s^2)(10,915~m)$ $P_{abs} = 1.10\times 10^8~Pa$ The water pressure at a depth of 10.915 km is $1.10\times 10^8~Pa$ (b) We can assume that the pressure inside the vessel is 1 atmosphere. Therefore, the pressure difference between the outside and inside of the hull is $1.10\times 10^8~Pa$. We can find the force on the top of the hull: $F = P~A = (1.10\times 10^8~N/m^2)(1.0~m^2) = 1.10\times 10^8~N$ The force exerted on the top of the hull due to the water pressure is $1.10\times 10^8~N$.