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