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Anni [7]
3 years ago
8

Change in v= 9.8 m/s2xt. The diagram shows a ball falling toward Earth in a vacuum.

Physics
2 answers:
Tresset [83]3 years ago
5 0

Answer:

Option A. 39.2 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 4 s

Final velocity (v) =?

v = u + gt

Since the initial velocity (u) is 0, the above equation becomes:

v = gt

Thus, inputting the value of g and t, we can obtain the value of v as shown below:

v = 9.8 × 4

v = 39.2 m/s

Therefore, the velocity of the ball at 4 s is 39.2 m/s.

aliya0001 [1]3 years ago
5 0

39.2 m/s

Explanation

a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s/s in The Physics Classroom Tutorial in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy.

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If you know how long it takes the clam to stop once it begins to hit the dirt,
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I would kick this question back to the Physics teacher.  It's meaningless,
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8 0
3 years ago
If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by
lina2011 [118]

Answer:

The value is  v = 1.309*10^{5}\ m/s

Explanation:

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The mass of the ordinary matter is M_{rod} =  1.90 *10^{41}\  kg

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Here G is the gravitational constant with a value

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Answer:

C.

Explanation:

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3 years ago
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