Answer:
[tex]a=20.14\ m/s^2[/tex]
Explanation:
The time period of the simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is the length of the pendulum
g is the acceleration due to gravity
We have,
T = 1.4 s, l = 1 m
So,
[tex]T^2=\dfrac{4\pi^2 l}{g}\\\\g= \dfrac{4\pi^2 l}{T^2}\\\\g= \dfrac{4\pi^2 \times 1}{(1.4)^2}\\\\g=20.14\ m/s^2[/tex]
So, the acceleration due to gravity of that planet is [tex]20.14\ m/s^2[/tex].
