Answer:
6.96 s
Explanation:
The period of a simple pendulum is given by:
[tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex]
where
L is the length of the pendulum and g the acceleration due to gravity.
In this problem, we have a pendulum with length L = 2.00 m, while the acceleration due to gravity is 1/6 that of the earth:
[tex]g' = \frac{g}{6}=\frac{9.8 m/s^2}{6}=1.63 m/s^2[/tex]
So, the period of the pendulum on the moon is
[tex]T=2 \pi \sqrt{\frac{2.00 m}{1.63 m/s^2}}=6.96 s[/tex]
