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The spring of the pressure gauge has a force constant of 2000N/m and the piston has a diameter of 3.00 cm .As the gauge is lowered into water, what change in depth causes the piston to move in by 0.55 cm?

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  1. The pressure needs to be enough to exert a force on the piston of
    F = kX = 2000*0.0055 = 11 N
    P = 11 N/(pi D^2/4) N/m^2
    (Make sure D in in meters)

    Set that equal to (depth)(density)*g and solve for depth

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