- #1
missashley
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A certain pendulum clock that works perfectly on Earth is taken to the moon, where g = 1.62 m/s^2. Acceleration of gravity is 9.81 m/s^2 on Earth. THe clock is started at 12:00:00 AM and runs for 22 h.
What will be the reading for the hours on the moon? answer in h
T = 1 second
g = 9.81
T = 2pi * square root of length/g or rearrange into g (T/2pi)^2 = l
l = 0.248
g = 1.63
T = 2pi * square root of l/1.63
T = 2.453
22h * 3600s = 79200 sec
79200/2.453 = 32286.99552
32286.99552/3600 = 8.9686 hours
i also tried
22*3600 = 79200s
79200 = 2pi * square root of l/9.81
l = 1557100708
T = 2pi *square root of 1557100708/1.63
T = 194197.8499s
194197.8499s / 3600 = 53.9438472 hr
What will be the reading for the hours on the moon? answer in h
T = 1 second
g = 9.81
T = 2pi * square root of length/g or rearrange into g (T/2pi)^2 = l
l = 0.248
g = 1.63
T = 2pi * square root of l/1.63
T = 2.453
22h * 3600s = 79200 sec
79200/2.453 = 32286.99552
32286.99552/3600 = 8.9686 hours
i also tried
22*3600 = 79200s
79200 = 2pi * square root of l/9.81
l = 1557100708
T = 2pi *square root of 1557100708/1.63
T = 194197.8499s
194197.8499s / 3600 = 53.9438472 hr
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