0
It would be extremely slow, and unsteady, and if it's too light, then it might not even work because it wouldn't have enough mass to interact with the moons gravity.
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
How much of the moon can you see if it's a full moon?
A whole moon it's that simple
Have space shuttles and current rockets taken humans as far as the moon in recent years?
A Space Shuttle has never taken a human to the moon.
What did man do on the moon in 1969?
It was a quick visit to the moon, samples of moon rocks were collected, photographs were taken, and brought back to Earth.
What are two types of moon craters?
Complex and Simple.
How do people live on the moon?
simple answer... THEY DON'T :D
Time period of simple pendulum is three seconds the same pendulum taken at the moon - what will be its new time period?
This pendulum, which is 2.24m in length, would have a period of 7.36 seconds on the moon.
Is a pendulum's time period increases or decreases when it taken from earth to moon?
Increases.
Can you use simple pendulum in moon?
Yes. The period of the pendulum (the time it takes it swing back and forth once) depends on the length of the pendulum, and also on how strong gravity is. The moon is much smaller and less massive than the earth, and as a result, gravity is considerably weaker. This would make the period of a pendulum longer on the moon than the period of the same pendulum would be on earth.
How would the time period of a simple pendulum clock be affected if it were on the moon instead of the earth?
The time period of a pendulum would increases it the pendulum were on the moon instead of the earth. The period of a simple pendulum is equal to 2*pi*√(L/g), where g is acceleration due to gravity. As gravity decreases, g decreases. Since the value of g would be smaller on the moon, the period of the pendulum would increase. The value of g on Earth is 9.8 m/s2, whereas the value of g on the moon is 1.624 m/s2. This makes the period of a pendulum on the moon about 2.47 times longer than the period would be on Earth.
What is the difference in period for a pendulum on earth and a pendulum on moon?
The period of a simple pendulum swinging at a small angle is approximately 2*pi*Sqrt(L/g), where L is the length of the pendulum, and g is acceleration due to gravity. Since gravity on the moon is approximately 1/6 of Earth's gravity, the period of a pendulum on the moon with the same length will be approximately 2.45 times of the same pendulum on the Earth (that's square root of 6).
What would the effect be on the period of the simple pendulum if the pendulum was moved from sea level to the top of a mountain or to the moon or to the sun?
As the force of gravity increases the period would decrease. So shortest period on the sun (if you can keep it intact), then sea level, then mountain top and then moon.
How would the period of a simple Pendulum be charged if the pendulum were moved from sea level to the moon?
The period is not likely to be charged. However, it would change due to the weaker gravitational force on the moon. Since the surface gravity of the moon is 0.165 that of the earth, the period would increase by a multiple of 1/sqrt(0.165) = 2.462 approx.
What is the best device when placed on the Moon would provide evidence of Moon rotation?
For earthbound investigators, the Foucault pendulum provides a nice proof.
Will a pendulum swing on the moon?
I think it will as it has mechanical parts to make the pendulum move, not 100% sure.
Does the period of a pendulum increase or decrease on Moon?
The period of a pendulum (for short swings) is about 2 PI (L/g)1/2. The gravity on the moon is less than that on Earth by a factor of six, so the period of the pendulum on the moon would be greater, i.e. slower, by about a factor of 2.5.
What is the time period of a pendulum on moon?
... dependent on the length of the pendulum. ... longer than the period of the same pendulum on Earth. Both of these are correct ways of finishing that sentence.
What is the periodic time of a 0.5m pendulum on the moon?
note: (g(moon)= 1/6g(earth))
