physics

What is the free-fall acceleration in a location
where the period of a 2.87 m long pendulum
is 3.4 s?
Answer in units of m/s
2

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asked by mike
  1. T^2 = 4pi^2(L/g) = 3.4^2
    39.48(2.87/g) = 11.56
    113.3/g = 11.56
    11.56g = 113.3
    g = 9.80 m/s^2

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    posted by Henry

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