Answer:
g = 12.22 m/s²
Explanation:
The time period of this pendulum is given as follows:
[tex]T = \frac{time\ taken}{no. of cycles}\\\\T = \frac{144\ s}{93}\\\\T = 1.55\ s[/tex]
but the formula for the time period of a simple pendulum is as follows:
[tex]T=2\pi \sqrt{\frac{l}{g}}\\[/tex]
where,
L = length of pednulum = 48 cm = 0.48 m
g = magnitude of th gravitational acceleration on this planet = ?
Therefore,
[tex]1.55\ s=2\pi \sqrt{\frac{0.48\ m}{g}}\\\\\sqrt{g} = 2\pi \sqrt{\frac{0.48\ m}{1.55\ s}}\\\\g = 3.49^{2}\\[/tex]
g = 12.22 m/s²
