After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cmcm. The explorer finds that the pendulum completes 103 full swing cycles in a time of 132 s. What is the value of the acceleration of gravity on this planet?

13.01 m/s²

Explanation:

The period of a simple pendulum is given as

T = 2π√ (L/g) ... Equation 1

Where T = Period of the simple pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet.

Given; T = 132/103 = 1.28 s, L = 54 cm = 0.54 m, π = 3.14

Substitute into equation 1

1.28 = (2*3.14) √ (0.54/g)

1.28 = 6.28√ (0.54/g)

Making g the subject of the equation,

√ (0.54/g) = 1.28/6.28

√ (0.54/g) = 0.2038

0.54/g = (0.2038) ²

0.54/g = 0.0415

g = 0.54/0.0415

g = 13.01 m/s²

Hence the value of the acceleration due to gravity on the planet = 13.01 m/s²