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11 June, 04:36

A pendulum is taken to another planet where it swings back and forth exactly 17 times every 35.66 seconds. The pendulum's arm is 0.74 m long. What is the acceleration due to gravity on this planet in units of m/s?

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  1. 11 June, 05:13
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    Answer: g = 6.65 m/s²

    Explanation: The frequency of a simple pendulum is given by the formulae below

    f = 1/2π * (√g/l)

    Where f = frequency = number of oscillation / time = 17/35.66 = 0.477Hz

    g = acceleration due gravity in another planet

    l = length of simple pendulum = 0.74m

    0.477 = 1/2π * (√g/0.74)

    0.477 = 1/2 * 3.142 * (√g/0.74)

    0.477 = 1 / 6.284 * (√g/0.74)

    0.477 = 0.1591 * (√g/0.74)

    0.477/0.1591 = (√g/0.74)

    2.998 = (√g/0.74)

    By taking the square of both sides

    8.988 = g / 0.74

    g = 6.65 m/s²
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