the period of a simple pendulum is 1s on earth when brought to a planet where g is one-tenth that of the earth its period become

A simple pendulum, with a fixed length and fixed mass at the end of the string, is set in motion on a planet that is 0.750 times the mass of the Earth and the same radius as that of the Earth. The period of this pendulum is what factor times its period on Earth (Tplanet = ____· T Earth)?

is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:

a) The period of a simple pendulum is T = 2𝜋L/g where L is the length of the pendulum and g is the acceleration due to gravity at the pendulum's location. Thus, if a pendulum has a period of T = 1.9 s on Earth where gEarth = 9.8 m/s2, its length isLEarth  =  gEarthT24𝜋2    =  m/s2 s 2 4𝜋2    =  m = cm

A simple pendulum has a period of 3.65 s on the surface of the Earth. If this pendulum were placed on the surface of the Moon, where the gravitational acceleration is 1.62 m/s2m/s2 , what would its period be?

A simple pendulum has a time period of  2.0sec at the earth's surface. It is taken to a height  Ro/2 above the earth's surface, where  Ro is the radius of the earth. What is the time period in sec?