What Is the Maximum Safe Depth for the Submarine with a 40cm Window?

  • Thread starter Thread starter KINkid92
  • Start date Start date
  • Tags Tags
    Confusion
AI Thread Summary
The discussion focuses on calculating the maximum safe depth for a research submarine with a 40 cm diameter window that can withstand a force of 1.20×10^6 N. The safe pressure is determined using the formula for pressure, leading to a calculated safe pressure of 9.55×10^6 N/m². The pressure at depth is derived from the sum of atmospheric pressure and the hydrostatic pressure from water, which involves the density of seawater and gravitational acceleration. Clarifications are sought regarding the pressure inside the submarine and its relationship to the external water pressure, particularly whether equal pressures would result in no force acting on the window. Understanding these pressure dynamics is crucial for determining the submarine's operational limits.
KINkid92
Messages
2
Reaction score
0

Homework Statement



A research submarine has a 40.0cm -diameter window 8.70cm thick. The manufacturer says the window can withstand forces up to 1.20×10^6N . What is the submarine's maximum safe depth?

Homework Equations



P= atmosheric pressure + ρg(depth)
Pressure = Force/crossectional area

The Attempt at a Solution



1.2*10^6N/∏(.4/2)^2= Safe pressure
9.55*10^6N/m^2=Psafe

Psafe= 1atm+ρgd=101300N+1025kg/m^3*9.8m^2
9.55*10^6kg/m^3/m^2=101300kg*m/s^2+ 1025kg/m^3*9.8m^2


I really just don't even know how to proceed figuring out this moshpit of units or even if I'm doing this right in the first place?
 
Physics news on Phys.org
What's the pressure inside the submarine and what is the difference with the total water pressure?
 
Inside the sub is 1 atm. And what do you mean? Difference between which values?
 
If the pressure inside is equal to the pressure outside - will there be any force on the window?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top