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jasenka [17]
3 years ago
9

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?
Physics
1 answer:
svetlana [45]3 years ago
4 0

Answer:

3.7 m/s^2

Explanation:

The period of a simple 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.

Calling L the length of the pendulum, we know that:

T_e = 2 \pi \sqrt{\frac{L}{g_e}}=1.50 s is the period of the pendulum on Earth, and g_e = 9.8 m/s^2 is the free-fall acceleration on Earth

T_m = 2 \pi \sqrt{\frac{L}{g_m}}=2.45 s is the period of the pendulum on Mars, and g_m = ? is the free-fall acceleration on Mars

Dividing the two expressions we get

\frac{T_e}{T_m}=\sqrt{\frac{g_m}{g_e}}

And re-arranging it we can find the value of the free-fall acceleration on Mars:

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

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A motorcycle going in a circle at a constant speed has changing velocity because the direction of the motion is constantly changing hence it is not in the state of motion.

Keywords: velocity, force, equilibrium, static equilibrium, dynamic equilibrium

Learn more about dynamic equilibrium from brainly.com/question/12880727

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4 years ago
Helpppppppppppp.......!
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Explanation:

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the C is correct one and surely I suggest you to go for it !
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