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6 December, 16:52

The atmospheric pressure above a swimming pool changes from 753 to 772 mm of mercury. the bottom of the pool is a rectangle (11 m cross product 21 m). by how much does the force on the bottom of the pool increase

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  1. 6 December, 19:40
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    Change in pressure : 772-753 = 19 mm Hg

    Since 1 mm Hg = 133 Pa Hg

    Then, 19 mm Hg = 2527 Pa Hg

    Area of rectangular pool = L * B = 11 x 21 = 231 m^2

    To find the force, we multiply the area by the change in pressure: 2527 (231) = 583737N
  2. 6 December, 20:45
    0
    Since we are talking about the pressure of water at a certain height, we can associate hydrostatic pressure or hydrostatic force. The formula relating this two is:

    P = F/A

    So,

    F = P*A

    To find the change of force, simply multiply the change in pressure to the cross-sectional area of the swimming pool. To report in terms of Newtons, let's convert mmHg to Pa (760 mmHg = 101325 Pa).

    F = (772 - 753 mmHg) (101325 Pa/760 mmHg) (11 m * 21 m)

    F = 585,151.875 N
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