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The pressure inside the submarine is maintained at 1.0 atm.

The sub is in salt water

- Thread starter jason_r
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- #1

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The pressure inside the submarine is maintained at 1.0 atm.

The sub is in salt water

- #2

alphysicist

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What have you tried so far?

The pressure inside the submarine is maintained at 1.0 atm.

The sub is in salt water

- #3

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I''ll show you what ive got so farHi jason_r,

What have you tried so far?

plz let me know if im doing this right, because i dont 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

- #4

alphysicist

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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?I''ll show you what ive got so far

plz let me know if im doing this right, because i dont have a key or anything.

thanks

P=P_o + pgh

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

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.

- #5

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ok, so..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.

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

- #6

alphysicist

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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.)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

- #7

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k thnxYes, 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.)

- #8

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

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

no seems to be replying..thanks

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