First solve the area of the ow the window os the submarine
A = pi* (0.15)^2= 0.0707 m2
The pressure it can with stand is
P= F/A = 1.2x10^6 N/ 0.0707 m2 = 16976527.26 Pa
Using the formala
P = pgh
Where p is the density of water
G is the accelartion due to gravity
H is the depth
H = P/pg
=16976527.26 Pa/ (1000 kg/m3)(9.81 m/s2)
H = 1730.53 m
