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How many swings of that same pendulum will it take for that process will it take on the moon? Physics! Help!?
Suppose the time for a chemical process to take place on earth is 10.0 swings of a certain standard pendulum.
If this process takes the same amount of time on our moon, where the objects fall toward its surface with an acceleration of 1.67 m s^-1, how many swings of the same pendulum will it take for that process on the moon?
1 Answer
- markumLv 51 decade agoFavorite Answer
For a pendulum, the period T = 2 * π * √ (L / g) where L is the length and g is the gravity. For earth, gE = 9.8m/s^2 and for the moon gM = 1.67 m/s^2 (I will assume the s^-1 in the question is a typo).
TE = 2 * π * √ (L / gE)
TM = 2 * π * √ (L / gM)
Rearranging the TM equation slightly
TM = 2 * π * √ (L / gE) * √ (gE / gM)
Substituting in TE
TM = TE * √ (gE / gM)
TM = 10s * √ (9.8 / 1.67)
TM = 24.2 seconds
Knowing the ratio of earth to moon gravity is enough to get the answer. The length L is not important
Source(s): http://en.wikipedia.org/wiki/Pendulum