Determine the length of a seconds pendulum at a place, where the acceleration due to gravity is `9.77 frac(m)(s^2)`

The length of seconds pendulum at a place wl}ere g = 4.9 m/ s^2 is

Find the change in length of a second 's pendulum , if the acceleration due to gravity at the place changes from 9.75 m/s^2 to 9.8 m/s^2 .

State whether the following statements are True of False,Correct the false statement . At the poles,the acceleration due to gravit is 9.77m//s^2 .

A body is projected vertically upwards with a velocity of 10 ms^-1 and another body is projected simultaneously from the same point with a velocity of 20 m s^-1 at an angle of pi/6 with the horizontal. The distance between the two bodies after one second from the time of projection is (Acceleration due to gravity is 10 ms^-2 )

When the length of a simple pendulum is increased by 36 cm, its time period of oscillation is found to increase by 25 %. The initial length of the simple pendulum 1s (Acceleration due to gravity= 9.8ms^(-2) )

If the density of the earth is tripled keeping its radius constant, then acceleration due to gravity will be (g=9.8 m/s^2)

The mass and weight of an object on Earth is 5 kg and 49 N respectively.What will be their values on the Moon? Assume that the acceleration due to gravity on the Moon is frac(1)(6) th of that on the Earth

The length of a seconds pendulum at a place where g = 9.8 frac(m)(s^2) is 0.994m If the amplitude is 0.15m, its maximum velocity is

If the density of the earth is doubled keeping its radius constant then acceleration due to gravity will be (g=9.8 m//s^(2))