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
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
square both sides
T^2=4PI^2 (l/g0
g= 4 PI^2 length/period^2
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