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The free-fall acceleration on the moon is 1.62 m/s2 . What is the length of a pendulum whose period on the moon matches the period of a 1.90-m-long pendulum on the earth
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The free-fall acceleration on the moon is 1.62 m/s2 . What is the length of a pendulum whose period on the moon matches the period of a 1.90-m-long pendulum on the earth

User Vigintas Labakojis
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1 Answer

1 vote

Answer:

P = 2 pi (L / g)^1/2

Let 1 represent earth and 2 the moon

P2^2 / P1 ^2 = L2 g1 / (L1 g2) dividing equations

L2 = (P2 / P1)^2 * g2 / g1 * L1

L2 = 1 * 1.62 / 9.80 * 1.9 = .314 m

User Ed McManus
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