Answer:
3.7 m/s^2
Explanation:
The period of a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where L is the length of the pendulum and g is the free-fall acceleration on the planet.
In this problem, we know that the period of the pendulum on Earth is:
[tex]T_e = 1.50 s[/tex]
while the period of the same pendulum on Mars is
[tex]T_m = 2.45 s[/tex]
And since the length of the pendulum L does not change, we can write:
[tex]\frac{T_e}{T_m}=\frac{2\pi \sqrt{\frac{L}{g_e}}}{2\pi \sqrt{\frac{L}{g_m}}}=\sqrt{\frac{g_m}{g_e}}[/tex]
where
[tex]g_e = 9.8 m/s^2[/tex] is the free-fall acceleration on Earth
[tex]g_m = ?[/tex] is the free-fall acceleration on Mars
Re-arranging the equation and substituting numbers, we find:
[tex]g_m = \frac{T_e^2}{T_m^2}g_e=\frac{(1.50 s)^2}{(2.45 s)^2}(9.8 m/s^2)=3.7 m/s^2[/tex]
