Gravitation: 4 Problems
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1) The following formula represents the period of a pendulum,T.
T = 2π(l/g)
(a) What would be the period of a 1.8 m long pendulum on the moon's surface? The moon's mass is 7.34 1022 kg, and its radius is 1.74x10^6 m.
2) A force of 40.9 N is required to pull a 10.2 kg wooden block at a constant velocity across a smooth glass surface on Earth. What force would be required to pull the same wooden block across the same glass surface on the planet Jupiter?
3) A 1.1 kg mass weighs 10.78 N on Earth's surface, and the radius of Earth is roughly 6.4 106 m. (Use G = 6.67 10-11 N·m2/kg2.)
(a) Calculate the mass of Earth.
kg
(b) Calculate the average density of Earth.
kg/m3
4) An apparatus like the one Cavendish used to find G has a large lead ball that is 6.2 kg in mass and a small one that is 0.040 kg. Their centers are separated by 0.055 m. Find the force of attraction between them. (Use G = 6.67 10-11 N·m2/kg2.)
N
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Solution Summary
4 problems related to gravitation and gravity are solved and explained in the solution.
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1) The following formula represents the period of a pendulum.
T = 2π(l/g)
(a) What would be the period of a 1.8 m long pendulum on the moon's surface? The moon's mass is 7.34 1022 kg, and its radius is 1.74x10^6 m.
The time period T of a simple pendulum of length l is given by
Here g is the acceleration due to gravity at the place of experiment.
Now the acceleration due to gravity at the surface of a planet is given by
Here G is universal gravitation constant, M is the mass of the planet and R is the radius of the planet.
Thus the acceleration due to gravity near the surface of moon will be given by the same formula.
Here mass of the moon M = 7.34*1022 kg
Radius of moon R = 1.74*106 m
And gravitation constant G = 6.67*10-11 N m2/kg2
Thus the acceleration at the surface of moon will be
m/s2
Thus the time period of the pendulum of length l = 1.80 m on the surface of moon is given by
Thus the period of small oscillation of pendulum will be 6.6 s.
2) A force of 40.9 N is ...
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