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Pressure on a window of a submarine

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data
    a research submarine has a 20 cm diameter window 8 cm think. the manufacturer says the window van withstand forces up to 1x106N. What is the submarine's maximum safe depth?


    2. Relevant equations
    p = po + [tex]\rho[/tex]gd


    3. The attempt at a solution
    p = F/A = 1X106 / 0.03 = 3.2x107 Pa

    so now do i use p = po + [tex]\rho[/tex]gd ??

    but if i do what would be my [tex]\rho[/tex]?
    [tex]\rho[/tex] = m/v = m/[tex]\pi[/tex]r2h
    how would i find mass?

    thanks in advance for help!
     
  2. jcsd
  3. Dec 2, 2009 #2
    ok so i solved it with the help of a different post but i had a question:
    she does p' = p1 + [tex]\rho[/tex]gd
    p' - p1 = [tex]\rho[/tex]gd

    and then she says that p1 = 1x105 Pa > atmospheric pressure. I was just wondering why?

    can someone explain how
    [tex]\Delta[/tex]p = p1 - po relates or is used in conjunction with p' = p1 + [tex]\rho[/tex]gd? is [tex]\Delta[/tex]p = p'
     
  4. Dec 2, 2009 #3
    anyone??
     
  5. Dec 2, 2009 #4
    i really just cant seem to wrap my head around pressure concepts, if someone knows how to explain it clearly or knows a website where i can see examples and stuff that would be great!
     
  6. Dec 3, 2009 #5

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    it gets confusing with so many different letter variables like [tex] p, p_1, p_o, p', \rho [/tex] (and Pa for Paschals to boot), so let's define the variables:

    let

    [tex]p_o [/tex] = atmospheric pressure = 1.01(10^5) Paschals
    [tex] p [/tex] = total pressure at depth d
    [tex] \rho [/tex] = density of water = 1000kg/m^3)

    Now when the top of the water is not sealed off (that is, it is exposed to the atmosphere),
    [tex]p = p_o + \rho gd[/tex], that is , the total pressure at depth d is the sum total of the atmospheric pressure and the water pressure at that depth.
     
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