An object is released from remainder on a planet that has actually no setting. The object falls easily for 3 m in the initially second. What is the magnitude of the acceleration due to gravity on the planet?
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The equation of activity that have the right to be offered to define the fall of this object is

`d = v_0t + at^2/2` .

Here, d is the distance that the object dropped in time t , and also a is the acceleration due to the gravity of the world. Since tbelow is...


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The equation of movement that have the right to be provided to explain the fall of this object is

`d = v_0t + at^2/2` .

Here, d is the distance that the object fell in time t, and also a is the acceleration as a result of the gravity of the planet. Because tbelow is no atmosphere, the gravity is the just force the object will certainly experience, so the gravity is the only pressure giving this acceleration.

Since the object in this case falls from rest, initial velocity `v_0 =0` . Then the equation activity becomes

`d = at^2/2` .

From below, the acceleration because of gravity have the right to be found:

`3 = a*1^2/2`

`a = 6 m/s^2`

The acceleration of the complimentary fall, or the acceleration as a result of gravity on this planet is 6 m/s^2.

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