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Ears popping when ascending or descending

  1. Apr 20, 2005 #1
    I have these problems about Fluids that I really need help with. Can you please tell me how to get started with these problems. Thanks alot!

    1) When u ascend or descend a great deal when driving in a car, your ears "pop," which means that the pressure behind the eardrum is being equalized to that outside. If this didn't happen, what would be the approximate force on an eardrum of area 0.50cm^2 if a change in altitude of 950m takes place?

    What I did was: Pressure= (density)(g)(height) Then (Pressure)(Area)= Force
    Is that right?
    I got .600045 for the force... which seems... wrong.

    2) During each heartbeat, approximately 70cm^3 of blood is pushed from the heart at an average pressure of 105 mm-Hg. Calculate the power output of the heart, in watts, assuming 70 beats per minute.

    for this one... i'm not sure of what to do... so plz help. thanks
     
  2. jcsd
  3. Apr 20, 2005 #2

    OlderDan

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    1) The height in the equation is the height of a column of a fluid assuming there is constant density. If you are using the density of air, then you should be getting the pressure difference between two levels, which is what you need. I don't know what your units are, or the density off hand, so I can't comment on the size of the force.

    2) The easiest approach is to model the exit from the heart as a cylinder of some cross section A, and think about how far the blood has to move. Then use force times distance to connect to work/energy.
     
    Last edited: Apr 20, 2005
  4. Apr 20, 2005 #3

    minger

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    If your answer is in Newtons, that's close to what I got :shrug:
     
  5. Apr 20, 2005 #4
    1. 950m seems like too much to consider the column as constant density. If you want a correct answer, assuming that the observer rises 950m from the sea level, use the barometric formula:

    [tex]p = p_0e^{-\frac{mgh}{KT}}[/tex]

    to derive the atmospheric pressure at the height of 950m.

    2. You know the pressure p. Then you have to manipulate the work formula:

    [tex]W = Fs = (pA)\frac{V}{A} = pV[/tex]

    where V is volume.

    [tex]P = \frac{W}{t} = p\frac{V}{t} = pR[/tex]

    where R is rate.

    From the given data of course you can find the rate of pumped blood in cube meters per second.
     
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