An ocean-going research submarine has a 40.0 cm -diameter window 9.00 cm thick. The manufacturer says the window can withstand forces up to 1.20×106 N . What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.
ρseawater = 1030 kg/m3
r = .2 m
g = 9.8 m/s2
p = F/A
p = p0 + ρseawatergd
A = πr2
The Attempt at a Solution
p0 = 1 atm = 101,300 Pa
p = (1.2*106 N)/(π*.22 m) = 9549296.586 Pa
9549296.586 Pa = 101300 Pa + (1030 kg/m3)(9.8 m/s2)d
d = (9549296.586 Pa - 101300 Pa)/((1030 kg/m3)(9.8 m/s2))
d = 936m
But the actual answer is 947m so I'm pretty sure I did something wrong or I'm missing something at some point because that is a significant difference. I'm pretty sure there's some point where I'm supposed to be using the volume of the window when I'm not but I'm not quite sure where I'm supposed to use it.