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21 April, 06:29

If the energy stored in the fully charged battery is used to lift the battery with 100-percent efficiency, what height is attained? Assume that the acceleration due to gravity is 9.8 m/s2m/s2 and is constant with height

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  1. 21 April, 09:45
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    h = 32059.37 m

    Explanation:

    Assuming the missing in formation as

    A certain lead-acid storage battery has a mass of 33 kg. Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.

    Now, Applying energy conservation (Electrical to potential)

    Electrical Energy E = I*V*t

    I = correct, V = voltage, t = time of flow of current

    E = 6*24*20*60*60.

    E = 10368 KJ

    Now this energy is used to lift the battery with 100% efficiency

    Hence,

    electrical energy E = potential energy P

    P = mgh

    m=mass of the battery, g = the acceleration due to gravity is 9.8 m/s^2

    h = height

    mgh = 10368 kJ

    33*9.8*h = 10368*1000

    h = 10368*1000 / (33*9.8)

    h = 32059.37 m
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