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## Homework Statement

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 1.20×10

^{6}N . What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.

ρ

_{seawater}= 1030 kg/m

^{3}

r = .2 m

g = 9.8 m/s

^{2}

## Homework Equations

p = F/A

p = p

_{0}+ ρ

_{seawater}gd

A = πr

^{2}

## The Attempt at a Solution

p

_{0}= 1 atm = 101,300 Pa

p = (1.2*10

^{6}N)/(π*.2

^{2}m) = 9549296.586 Pa

9549296.586 Pa = 101300 Pa + (1030 kg/m

^{3})(9.8 m/s

^{2})d

d = (9549296.586 Pa - 101300 Pa)/((1030 kg/m

^{3})(9.8 m/s

^{2}))

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.