Answer:
3.7 m/s^2
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 is the free-fall acceleration on the planet.
Calling L the length of the pendulum, we know that:
[tex]T_e = 2 \pi \sqrt{\frac{L}{g_e}}=1.50 s[/tex] is the period of the pendulum on Earth, and [tex]g_e = 9.8 m/s^2[/tex] is the free-fall acceleration on Earth
[tex]T_m = 2 \pi \sqrt{\frac{L}{g_m}}=2.45 s[/tex] is the period of the pendulum on Mars, and [tex]g_m = ?[/tex] is the free-fall acceleration on Mars
Dividing the two expressions we get
[tex]\frac{T_e}{T_m}=\sqrt{\frac{g_m}{g_e}}[/tex]
And re-arranging it we can find the value of the free-fall acceleration on Mars:
[tex]g_m = g_e \frac{T_e^2}{T_m^2}=(9.8 m/s^2)\frac{(1.50 s)^2}{(2.45 s)^2}=3.7 m/s^2[/tex]
