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dedylja [7]
2 years ago
11

If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gr

avity at the surface would become.
Physics
1 answer:
timofeeve [1]2 years ago
7 0

When the mass of Earth doubles, acceleration due to gravity doubles as well.

<h3>Acceleration due to gravity</h3>

Apply Newton's second law of motion;

F = mg --- (1)

where;

  • m is mass of the object
  • g is acceleration due to gravity

F = GmM/R²  --- (2)

where;

  • M is mass of Earth
  • R is radius of Earth

Solve (1) and (2)

mg =  GmM/R²

g = GM/R²

when the mass of Earth doubles, acceleration due to gravity becomes;

g' = G(2M)/R²

g' = 2(GM/R²)

g' = 2g

Thus, when the mass of Earth doubles, acceleration due to gravity doubles as well.

Learn more about acceleration due to gravity here: brainly.com/question/88039

#SPJ1

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qaws [65]

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

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II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

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From the graph given above, the car has a constant velocity between B and C.

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6 0
3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
3 years ago
A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?
Kazeer [188]
-- She went up for 0.4 sec and down for 0.4 sec.

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3 years ago
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oksian1 [2.3K]

Answer:

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4 0
3 years ago
What is the most appropriate tool to use when collecting data from the investigation? List, table, graph or chart?
forsale [732]
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3 years ago
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