Asked by Angel

What is the free-fall acceleration in a location
where the period of a 0.594 m long pendulum
is 1.55 s s?
Answer in units of m/s

Answers

Answered by bobpursley
period=2PI*sqrt(l/g)

square both sides

T^2=4PI^2 (l/g0

g= 4 PI^2 length/period^2

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