Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 52.0 cm. The explorer finds that the pendulum completes 91.0 full swing cycles in a time of 140 s. What is the value of the acceleration of gravity on this planet?

Answer 1

Answer:

8.67340834769 m/s²

Explanation:

Time period is given by

$T=\dfrac{140}{91}\\\Rightarrow T=1.53846153846$

L = Length of pendulum = 52 cm

g = Acceleration of gravity on the planet

Time period also is given by

$T=2\pi\sqrt{\dfrac{L}{g}}\\\Rightarrow g=4\pi^2\dfrac{L}{T^2}\\\Rightarrow g=4\pi^2\dfrac{52\times 10^{-2}}{1.53846153846^2}\\\Rightarrow g=8.67340834769\ m/s^2$

The value of the acceleration of gravity on this planet is 8.67340834769 m/s²