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Pressure on a lungfish?

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm having some trouble with the concept of pressure, so I'd like to make sure I'm not doing this wrong!

    What is the pressure 45 m underwater?

    2. Relevant equations
    P = Patm + ρgh

    3. The attempt at a solution
    Density of seawater = 1020 kg/m3
    Patm = 101300 Pa

    So P = 101300 Pa + (1020 kg/m3)(9.81 m/s2)(45 m) = 154 274 Pa or 154 kPa.

    My question is - the pressure on something huge should be the same as the pressure on something small, right?

    Also, I read something about the pressure on a diver being the atmospheric pressure plus the water pressure. But the atmospheric pressure is taken into account already by the formula used, so I don't need to add it in again, correct?

    Thanks :)
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 4, 2010 #2


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    Right basically. So the question really doesn't have much to do with lungfish. It's just what the pressure is 45m below the sea. But I don't see why you think 101300 Pa + (1020 kg/m3)(9.81 m/s2)(45 m) = 154 274 Pa. That's WAY off. Can you check it again?
  4. Apr 4, 2010 #3
    Whoa, you're right. I don't know how I got that...

    (1020 kg/m^3)(9.81m/s^2)(45 m) = 450 279 Pa

    Add that to 101 300 Pa and you get 551 579 Pa. I must have skipped a button on my calculator the first time.

    I was just confused because the question says "estimate", but this isn't really an estimate. Usually my prof would only ask us to estimate if we didn't exactly know one of the values to use.
  5. Apr 4, 2010 #4


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    Guess I'm not sure what "estimate" means in this context either. Unless it just means not using the calculator at all and rounding things off so you get 100000+1000*10*45=550000. Which is pretty close. And doing that might have helped you to realize your were missing a button somehow.
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