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    A swimming pool is 20m by 30m and contains water to a depth of 2m. What is the change in GPE of the water when the pool is drained?(Ignore any further changes after the water leaves the pool and assume it is drained by pulling a plug from the bottom of the pool. Density of water = 1000kgm^-3)

    I have attempted to do this question and I found the volume which is 1200m^3
    I found the mass by using p x v = m and it came out as 1.2 x 10^6. I then used the GPE equation where GPE=mgh.

    Subbing in I get 1.2 x 10^6 x 9.8 x 2.
    I get the answer 23520000.
    However they get the answer 5.88 MJ which I do get if I do
    1.2 x 10^6 x 9.8 x 0.5.

    Why is the height counted as 0.5 not 2. Please explain
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    We are going to calculate GPE of the pool, having in mind that the center of mass of the pool is located at a point h/2.
    Thus, our formula GPE = mgh now is GPE = mg*(h/2).
    We are going to calculate m. Density of water is 1000 kg/m^3.
    m = 1000 kg/m^3 * 20 m * 30 m * 2 ( 2 m is our h for the calculation of mass, as we need to include all of the mass of water). m = 1,2 * 10 ^ 6.
    GPE = (1,2 * 9.8 * 10 ^ 6)/2 = 5,88 * 10 ^ 6.(" ^ " =" to the power of")
 
 
 
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