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Density/Sea Water Problem

  1. Dec 2, 2007 #1
    [SOLVED] Density/Sea Water Problem

    Assume the density of sea water is 1024 kg/m^3 and take the atmospheric pressure equal to 101000Pa. The acceration due to gravity is 9.8m/s^2. Calculate the pressure at an ocean depth of 1260m. Answer in Pa.

    Calculate the total force exerted on the outside of a circular submarine window of a diameter of 27.2cm at this depth. Answer in units of N.




    p=m/v | P=F/A | P=pgh
    p=density, P= pressure, F=force, A=area, g=gravity, h=height




    For this problem, I used P=pgh without regard to the pascals. Then for part II, I used my pressure from the first part and plugged it into P=F/A... I used the area of the submarine window with a diameter of 27.2. Neither got me the right answer.... and I'm not sure what equations to use now.
     
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  3. Dec 2, 2007 #2

    Kurdt

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    Did you add the the pressure due to the atmosphere to the pressure you calculated due to the water?
     
  4. Dec 2, 2007 #3
    No, thank you! Although I'm not sure why you would have to add the pascals.... I thought the equation P=pgh already encompassed the atmospheric pressure ><
     
  5. Dec 2, 2007 #4

    Kurdt

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    The way we calculate pressure is to find the weight of whatever substance it is acting on a certain area. if you look at the expression you have it should be obvious now that it only takes into account water pressure, thus you have to add on the atmospheric pressure as well.
     
  6. Dec 2, 2007 #5
    Well that makes sense. See our teacher never explains to us these things.

    However, now when I use the pressure that I got, 12644352 and plug it into P=F/A for part II, I get the wrong newtons.
     
  7. Dec 2, 2007 #6

    Kurdt

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    What exactly are the calculations you're making?
     
  8. Dec 2, 2007 #7
    I got the area of the submarine window... (27.2cm/2)^2 and pressure: P using F=PxA.
     
  9. Dec 2, 2007 #8

    Kurdt

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    be careful of the units then because the window size is given in cm and the SI units of pressure are newtons per meter square.

    EDIT: Just noticed the window is circular. What is the area of a circle?
     
    Last edited: Dec 2, 2007
  10. Dec 2, 2007 #9
    I got the area to be .018496m^2 >< but it doesn't work! arghh...
     
  11. Dec 2, 2007 #10

    Kurdt

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    I get a different area. Remember that the area of a circle is [itex] \pi r^2[/itex] where [itex] r = d/2 [/itex].
     
  12. Dec 2, 2007 #11
    Oh wow, thank you so much. I hate overlooking the obvious... it's been a long week. ><
     
  13. Dec 2, 2007 #12

    Kurdt

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    No problem. We've all made the same mistakes at some point.
     
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