Physics: After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . the explorer finds that the pendulum completes 99.0 full swing cycles in a time of 128 s . what is the magnitude of the gravitational acceleration on this planet? express your answer in meters per second per second.

It takes 3 secondsExplanation:...

After landing on an unfamiliar planet, a space explorer

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12.03.2022

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

$T = \frac{t}{n}$

$T = \frac{128}{99}$

T = 1.3 sec

Time period of the pendulum is also given as

$T = 2\pi \sqrt{\frac{L}{g}}$

$1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}$

g = 12.4 m/s²

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