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Yesterday, 11:16

When a submarine dives to a depth of 5.0 * 102 m, how much pressure, in pa, must its hull be able to withstand? how many times larger is this pressure than the pressure at the surface?

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  1. Yesterday, 11:38
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    Pressure is force over surface.

    p=F/S.

    In this case the force is given by the weight of the water G, so

    p = G/S;

    G=m*g;

    m=ρ*V,

    V=S*h

    since we are interested only in the volume of water which presses on the surface of the submarine S, at depth h.

    So:

    p = ρ*S*h*g/S = ρ*g*h.

    This is the pressure at a certain depth h, exerted only by the water so we must add the atmospheric pressure p0.

    Therefore the total pressure is pt = p0 + ρ*g*h.

    We can see that when h=0, pt=p0 which is correct since at the surface we have only atmospheric pressure.

    The ratio pt/p0 = 1 + ρ*g*h/p0

    p0=101325 Pa

    h = 500 m

    g = 9.8 m/s^2

    ρ = 1.025*10^3 kg/m^3.

    We substitute in the equation and get pt/p0 = 50.56.

    So the pressure at 500m depth is 50.56 times greater than at the surface.
  2. Yesterday, 11:45
    0
    Depth = 5.0 Ă - 10^2 m

    Density of sea water = 1.025 x 10^3

    Pd = Po + pgh

    Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa

    Since the normal pressure is retained in the hull, no need to bother about Po Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5

    So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56

    So it is 49.56 times larger.
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