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trasher [3.6K]
3 years ago
13

A basesball is dropped from 100 meters above the surface of the earth. The baseball falls to the ground. What would happen if th

is baseball was dropped from 100 meters above the surface of the moon?
Physics
2 answers:
Readme [11.4K]3 years ago
6 0

Answer:

It would fall toward the Moon's surface

Explanation:

As you already know, a ball that is thrown 100 meters above the surface of the earth, would fall on the surface of the earth, due to the gravity of our planet, which attracts any object around it to its surface (and that is what we keep them attached to the planet's floor). The same would happen if a ball were launched 100 meters above the surface of the moon, but the ball would fall on the surface of the moon, for the same reason that is the gravity of the moon. As the Moon is much smaller than the Earth, its force of gravity is also less, therefore the bodies are attracted with less force to the lunar surface, even so the ball would fall on the surface of the moon because the distance between the ball and the earth is much greater than the distance between the moon and the ball.

Serjik [45]3 years ago
4 0

Given that, A basesball is dropped from 100 meters above the surface of the earth. If the same baseball was dropped from 100 meters above the surface of the moon, it will take more time to hit the ground as compare to the ball dropped on earth. This is moon's gravity is one-sixth of that of earth. Object falling on earth possess more force of attraction. So it will reach the earth in lesser time. At moon the force of attraction is low compare to that of earth. Object will take more time to reach the surface.

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Explanation:

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1. What could scientists learn by using the Law of Superposition?
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3 years ago
An object moves at 60 m/s in the +x direction. As it passes through the origin it gets a 4.5 m/s^2 acceleration in the -x direct
Brut [27]

Answer:

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Explanation:

a) Let the starting position be origin and time be t.

  After time t displacement, s = 0 m

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

 We have equation of motion s = ut + 0.5 at²

 Substituting

        s = ut + 0.5 at²

        0 = 60 x t + 0.5 x (-4.5) x t²        

        2.25t² - 60 t = 0

        t² - 26.67 t = 0

        t (t-26.67) = 0

      t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

  Time , t = 26.67

 Substituting

        v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction

 

4 0
3 years ago
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