# Fluids at Rest!

1. Feb 21, 2005

### cdhotfire

While exploring a sunken ocean liner, the principal researcher found hte absolute pressure on the robot observatino submarine at the level of the ship to be about 413 atmospheres. The density of seawater is $1025-kg/m^3$.

Calculate the gauge pressure $p_g$ on the sunken ocean liner.

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This is the problem, i have the formula:
$p = p_{o} + p(to-lazy-to-look-up-density)gh$
I do not know which $p$ or $p_o$ but for the matter of making my point lets say, $p_{o} = 413 at$ or $41836900-Pa$. So
then:
$p = 41836900-Pa + (1025-kg/m^{3})(9.81-m/s^{2})h$
So Ive got $p$ and $h$ wish means, I have nothing. :uhh: .

Any ideas, would be appreciated.

2. Feb 22, 2005

### FredGarvin

Do you understand the relationship between gauge pressure and absolute pressure? This problem is a lot easier than you are making it out to be.

3. Feb 22, 2005

### minger

Yes, remember that absolute pressure is atmospheric pressure plus guage. The only information you will need for this problem is depth....and actually you could live with just depth and temperature, but depth and density/specific weight will be even easier.

4. Feb 22, 2005

### HallsofIvy

Staff Emeritus
Actually, you don't need that because you are given the absolute pressure in atmosperes: guage pressure is absolute pressure minus atmospheric pressure and atmospheric pressure is "1 atomosphere" so ....

IF you are asked to deduce the depth of the wreck from that, then you will need to know the density of water. I presume that's the next question!

5. Feb 22, 2005

### FredGarvin

Just re-read the last post so I deleted.

Last edited: Feb 22, 2005
6. Feb 22, 2005

### minger

Actually, yea, I didn't even read the initial problem. Yes, do what HallsofIvy said. Absolute = Atmospheric + Guage, and I hope you can figure out in atmospheres, what atmospheric prsesure is.