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Please help me fluids

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data
    In scuba diving, a regulator is used so that the pressure of the air the diver breathes is close to that of the ambient water. A reckless swimmer decides to use a hose sticking out of the surface to breathe underwater while diving in a lake. When the air pressure in the lungs is at a pressure of around 0.130 atm below the ambient pressure, lung injury may occur. Find the depth at which the swimmer would experience such a pressure differential.

    2. Relevant equations
    p = po +pgd



    3. The attempt at a solution
    p = po + pgd
    -po/pg = d
    -101300 / ((0.13 * 101300) * 9.8) = 0.78

    What am I doing wrong??
     
    Last edited: Nov 16, 2007
  2. jcsd
  3. Nov 16, 2007 #2
    Oh wait I see something I completely missed!

    p = 1 atm - 0.13 atm = 0.87 atm * 101300 pa = 88131 pa
    p - po / pg = d
     
  4. Nov 16, 2007 #3

    mgb_phys

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    Science Advisor
    Homework Helper

    pressure = force / area.
    force = m g h
    I 1m^3 of sea water has a mass of 1025kg then 1 meter column of water exerts 1025kg / m^2
    F = 1025 * 9.8 * h = 10045N * h
    Atmosphere = 101.4KPa = 101400 N /m^2

    0.13 * 101400 = 10045 h
    h = 1.3m

    As a quick check, if atmsopheric pressure is 101.4KPa this is about 10Kg/m^2 or 1 atmosphere is 10m of water.
     
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