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How many swings of the same pendulum will it take for that process on the moon?
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^2, how many swings of the same pendulum will it take for that process on the moon?
Please help! Thanks :)
1 Answer
- Violet WLv 71 decade agoFavorite Answer
The period of a pendulum's swing is approximately given by:
T=2pi x sqrt(L/g)
L = length, g = local gravity.
On the earth, ge = 9.81 m/s^2
On the moon, gm = 1.67m/s^2
Since T is proportional to sqrt(1/g), we have:
Tm/Te = sqrt (ge/gm)
or
Tm = sqrt(ge/gm) x Te
= sqrt(9.81/1.67) x Te
Tm = 2.4237 x Te
On the moon, each swing of the pendulum takes 2.42 times as long as a swing on the earth. That means there will be fewer swings in the same amount of time.
N = 10.0 swings / 2.4237 = 4.13 swings
Answer: 4.13 swings on the moon.
