What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a depth of 9400 m in sea water? the density of sea water is 1025 kg/m3?

Calculate the pressure due to sea water as density*depth.

That is,

pressure = (1025 kg/m^3) * ((9400 m) * (9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is 101.3 kPa

Total pressure is 94423 + 101.3 = 94524 kPa (approx)

The area of the window is π (0.44 m) ^2 = 0.6082 m^2

The force on the window is

(94524 kPa) * (0.6082 m^2) = 57489.7 kN = 57.5 MN approx