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# Answer to Question #22019 in Astronomy | Astrophysics for somnath chatterjee

Question #22019
1) a seconds pendulum is taken to the moon where the value of g is a sixth of the value on earth. the time period of the pendulum will be------

a) 1sec b) 2 sec c) 5sec d) none of these

2)if a force of 1 newton acts on a mass of 500 gms, acceleration will be?
1
Expert's answer
2013-01-17T09:16:38-0500
1)
T = 2*pi*Sqrt[l/g]
T -& the time period of the pendulum
g - acceleration due to gravity
l - the length of the pendulum
T1 = 2*pi*Sqrt[l/g] - on the Earth
T2 = 2*pi*Sqrt[l/(g/6)] - on the Moon
Therefore
T2 = T1 * Sqrt[6]
Answer: d) none of these
2)
F = m*a
F - force
m - mass
a - acceleration
a = F/m = 1H/500g = 1H /0.5 kg = 2 m/s^2

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