What is the Maximum Safe Depth for a Submarine Window?

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The discussion centers on calculating the maximum safe depth for a submarine window with a diameter of 20cm and a thickness of 8cm, capable of withstanding forces up to 10^6N. The pressure exerted on the window is derived from the equation P = P(atmosphere) + ρ*g*h, where the internal pressure is maintained at 1 atm. Calculations show that the maximum pressure the window can handle is approximately 3.18*10^7 Pa, leading to a maximum depth of around 3153m. The importance of considering the internal pressure when calculating the external pressure on the window is emphasized. Ultimately, the correct approach confirms the maximum safe depth for the submarine window is approximately 3153m.
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1. A research submarine has a 20cm diameter window, 8cm thick. The manufacturer says the window can withstand forces up to 10^6N. What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1atm.



2. Homework Equations P = P(atmosphere) + ρ*g*h


3. I tried finding the pressure on the window, using P = F/A and got the maximum pressure the submarine window can handle as P = 10^6/(pi*(0.1)^2) = 3.18*10^7 Pa.
I then used P = P(atmosphere) + ρ*g*h
and substitued, P as 3.18*10^7, P(atmosphere) as 101,300, ρ(used density of seawater) as 1030, g as 9.8 and found h to be approximately 3140m.
The solutions found h to be 3153m.
I don't know where I have gone wrong...
 
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Did you take into account that the internal pressure is 1 atm?
 
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Thanks I didn't consider it before but thinking about it in force perspective so it made more sense. I did Net force = Force ourtside window - Force inside window. Therefore 10^6 = Fout - Fin. Then Fout = P(outside)*A and Fin = P(inside)*A. 10^6/A = P(outside) - P(inside). 10^6/A = ρ*g*h + P(atmosphere) - P(inside submarine). Now P(atmosphere) = 1atm=P(inside submarine), ρ.=1030 Therefore 10^6/A = ρ*g*h.
Then I h to be (10^6)/(pi*(0.1^2)*9.8*1030) = 3153m.
THANK YOU VERY MUCH!
 
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