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27 May, 19:27

What pressure, in atmospheres, is exerted on the body of a diver if he is 38 ft below the surface of the water when atmospheric pressure at the surface is 750 mmHg? Assume that the density of the water is 1.00g/cm3=1.00*103kg/m3. The gravitational constant is 9.81m/s2, and 1Pa=1kg/m-s2.

Express your answer using two significant figures.

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  1. 27 May, 19:34
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    The pressure of diver = atmospheric pressure + water pressure

    atmospheric pressure = 750 mmHg (as given) = 750 / 760 atm = 0.987 atm

    Water pressure is

    P = hρg

    where

    h = height of water = 38 ft

    1 ft = 0.3048

    38 ft = 11.58 m

    ρ = density = 1000 Kg / m³

    g = gravitational constant = 9.81 m/s2

    P = 11.58 X 1000 X 9.81 = 113599.8 Kg / m s^2 Or N / m^2

    1 N / m^2 = 1 pa = 9.869 X 10^-6 atm

    P = 113599.8 Pa = 1.12 atm

    Total pressure = 1.12 + 0.987 atm = 2.107 atm = 2.1 atm (two significant figures)
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