A research submarine has a 40.0 cm-diameter window 8.50 cm thick. The manufacturer says the window can withstand forces up to 1.20×10^6 N. What is the submarine's maximum safe depth?
The Attempt at a Solution
The area which force will be exerted on is
A= Pai * r^2 = 0.5027 m^2
Maximum pressure allowed
p' = 1.20×10^6 N / 0.5027 m^2 = 2387109.608 Pa = 23.559 atm
Maximum pressure allowed from outside
p = 23.559 - 1 = 22.559 atm = 2285784.608 Pa
h =p/rou(density) * g = 2285784.608 / (1030 * 9.8) = 226.449 m
Which is not right.
I have two questions,
1, what is the 0.085m thickness given for
2, Was it correct that I have used density of sea water instead of water since it is a submarine?