After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 47.0 cm . The explorer finds that the pendulum completes 91.0 full swing cycles in a time of 125 s. What is the magnitude of the gravitational acceleration on this planet?

Explanation:

Given that,

The length of a simple pendulum, l = 47 cm = 0.47 m

The pendulum completes 91.0 full swing cycles in a time of 125 s.

The time period of a simple pendulum is given by :

Also,

t is 125 s and n = 91 cycles

Substitute all the values to find g.

So, the gravitational acceleration on this planet is .

answer: the answer is a

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