What is the submarine's maximum safe depth?

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The discussion centers on calculating the maximum safe depth for a research submarine based on the specifications of its window. The window, with a diameter of 17 cm and thickness of 6.0 cm, can withstand a force of 1.0X10^6 N. The pressure inside the submarine is maintained at 1.0 atm, while seawater has a density of 1030 kg/m^3. The equation used to find depth is p = p0 + pgd, but the initial attempt at solving it resulted in an incorrect depth of 89.033 m. The relationship between force and pressure is highlighted, indicating the need to consider area in calculations.
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Homework Statement



A research submarine has a 17cm diameter window 6.0 cm thick. The manufacturer says the window can withstand forces up to 1.0X10^6 N. What is the submarine's maximum safe depth? Assume that the pressure inside the submarine is maintained at 1.0atm and that the density of seawater is 1030 kg/m^3.

Homework Equations



p = p0 + pgd

The Attempt at a Solution



I tried the following

1.0X10^6 = 1.013X10^5 + (1030)(9.8)d, and I got d = 89.033 and of course it was wrong.
 
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What's the relationship between force and pressure ?
 
p = F/A
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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