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At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is $1.63 \mathrm{m} / \mathrm{s}^{2},$ if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock's hour hand to make one revolution on the Moon.

$T_{2}=2.452 \mathrm{h}$

Physics 101 Mechanics

Physics 103

Chapter 16

Oscillatory Motion and Waves

Periodic Motion

Wave Optics

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So here we have the pendulum clock on the moon and on Earth. We know that the period of a pendulum is equaling two pi multiplied by the length of by the by the square root of the length of the pendulum divided by the acceleration due to gravity. And so we can say that deep the time period of the pendulum is inversely proportional to the square root of G. And so we can then say that the time period on the moon divided by the time on earth, the period on on Earth rather would be equaling the square roots of the acceleration due to gravity on Earth divided by the acceleration of gravity on the moon. And so we can then say that the time period on the moon would be equaling. The time period on the earth multiplied by 9.80 meters per second squared divided by 1.63 meters per second squared. And this is equaling 2.45 times the period on earth. So essentially it'll take. We can say that. Then the question is asking us. Find the time and ours it takes for the clock's hour and to make one revolution on the moon. Now we know that, uh, on earth for the hour hand, it takes 12 hours. So 12 hours to make one revolution on, huh? So, essentially, we have 12 hours times 2.45 That factor. And we can say that Ah, 29 0.4 hours for our hand to make one evolution on the moon. And so, essentially, this would be our final answer. That is the end of the solution. Thank you for watching.

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