Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 48.0 cm . The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 141 s.

Required:
What is the value of g on this planet?

Answer 1

8.24 m/s²

Step-by-step explanation:

Applying,

T = 2π√(L/g).................... Equation 1

Where T = period of the pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet, π = pi

From the question, we were asked to find the value of g,

There we make g the subject of the equation

g = 4π²L/T²..................... Equation 2

Given: L= 48 cm = 0.48 m, T = (141/93) s = 1.516 s

Constant: π = 3.14

Substitute these values into equation 2

g = 4(3.14²)(0.48)/(1.516²)

g = 18.93/2.298

g = 8.24 m/s².

Hence the acceleration of the planet is 8.24 m/s²