Question

A research submarine has a 40-cm diameter window that is 8.0 cm thick. The manufacturer says the window can withstand forces up to 1.0 times 10^6 N. Part A What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm.

Answer 1

The thickness is irrelevant, that gets its way into the strength as reported by the manufacturer.

What is missing is the density of salt water. Let's denote it ρ.
The pressure at depth d then is:
p = p0 + ρ g d
, where p0 is the atmospheric pressure at the surface of the sea. We assume it is equal to the 1 atm inside the sub.

Hence the pressure difference between inside and outside of the window is:

Δp = ρ g d

The force on the window is this pressure difference, multiplied by the area of the circular window of diameter D. The latter equals
A = ¼ π D².

So
F = p A = ¼ π D ²ρ g d.
This should be less than Fmax.

Hence d < Fmax / (¼ π D ²ρ g ).

You have all the numbers except the density of sea water. Look it up in a table or on the net and plug it in along with the other numbers.

Density of salt water approximately = 1030 kg/m^3

d = 1*10⁶ / ((1/4)*0.40² *1030*9.8 ) = 2476.71 m