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liubo4ka [24]
3 years ago
12

Astronauts on the first trip to Mars take along a pendulumthat has a period on earth of 1.50 {\rm s}. The period on Mars turns o

ut to be 2.45{\rm s}.What is the free-fall acceleration onMars?gmars= m/s2
Physics
1 answer:
Fynjy0 [20]3 years ago
7 0

Answer:

3.7 m/s^2

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

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:

T_e = 1.50 s

while the period of the same pendulum on Mars is

T_m = 2.45 s

And since the length of the pendulum L does not change, we can write:

\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}}

where

g_e = 9.8 m/s^2 is the free-fall acceleration on Earth

g_m = ? is the free-fall acceleration on Mars

Re-arranging the equation and substituting numbers, we find:

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

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