Answer :
Answer:
[tex]g=9.83\ m/s^2[/tex]
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 :
[tex]T=2\pi \sqrt{\dfrac{L}{g}} \\\\or\\\\g=\dfrac{4\pi^2L}{T^2}[/tex]
Also,
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{t}{n}[/tex]
t is 125 s and n = 91 cycles
Substitute all the values to find g.
[tex]g=\dfrac{4\pi^2\times 0.47}{(\dfrac{125}{91})^2}\\\\g=9.83\ m/s^2[/tex]
So, the gravitational acceleration on this planet is [tex]9.83\ m/s^2[/tex].