What Is the Maximum Safe Depth for a Submarine with a 20 cm Window?

[jason_r]

A research submarine has a 20.0 cm-diameter window 8.90 cm thick. The manufacturer says the window can withstand forces up to 1.10×10^6 N. What is the submarine's maximum safe depth?

The pressure inside the submarine is maintained at 1.0 atm.
The sub is in salt water

 

[alphysicist]

Hi jason_r,

A research submarine has a 20.0 cm-diameter window 8.90 cm thick. The manufacturer says the window can withstand forces up to 1.10×10^6 N. What is the submarine's maximum safe depth?

The pressure inside the submarine is maintained at 1.0 atm.
The sub is in salt water

What have you tried so far?

 

[jason_r]

Hi jason_r,



What have you tried so far?

I''ll show you what I've got so far
please let me know if I am doing this right, because i don't have a key or anything.
thanks

P=P_o + pgh
1.10 x 10^6= 1 + (1030)(9.8)(h)
h=1.1 x 10^3m

 

[alphysicist]

I''ll show you what I've got so far
please let me know if I am doing this right, because i don't have a key or anything.
thanks

P=P_o + pgh
1.10 x 10^6= 1 + (1030)(9.8)(h)

There are two problems with this line. You have 1.10 x 10^6 on the left side; that is the force the window experiences, but in this equation you need the pressure that's on the window. So how can you find the pressure on the window if you know the force on the window?

Also, the first term on the right side: you seem to be saying that P_o is 1 atm, which is definitely true; however, to match the other term you need P_o to be in units of Pascals.

 

[jason_r]

There are two problems with this line. You have 1.10 x 10^6 on the left side; that is the force the window experiences, but in this equation you need the pressure that's on the window. So how can you find the pressure on the window if you know the force on the window?

Also, the first term on the right side: you seem to be saying that P_o is 1 atm, which is definitely true; however, to match the other term you need P_o to be in units of Pascals.

ok, so..
this is what i have now:

P=P_o + pgh
35014087.48=(P_o(air) - P_o(sub)) + (1030)(9.8)(h)
h=3.47 x 10^3
(does the pressure of air on top of the sea cancel the pressure of air in the sub?)
Also can you confirm that i have the correct anser?
thanks

 

[alphysicist]

ok, so..
this is what i have now:

P=P_o + pgh
35014087.48=(P_o(air) - P_o(sub)) + (1030)(9.8)(h)
h=3.47 x 10^3
(does the pressure of air on top of the sea cancel the pressure of air in the sub?)
Also can you confirm that i have the correct anser?
thanks

Yes, those pressures will cancel if you calculate it in one step, and that answer looks right to me. (As long as you have the right value for the density of seawater.)

 

[jason_r]

Yes, those pressures will cancel if you calculate it in one step, and that answer looks right to me. (As long as you have the right value for the density of seawater.)

k thnx

 

[jason_r]

btw can you help me on these questions:

https://www.physicsforums.com/showthread.php?p=1873415#post1873415
https://www.physicsforums.com/showthread.php?p=1873418#post1873418

no seems to be replying..thanks