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3 years ago

When the distance between two masses is doubled, the gravitational attraction between

Physics

1 answer:

mrs_skeptik [129]3 years ago

5 0

The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

$r'=2r$

Therefore, the new force will be

$F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}$

Therefore, the new force will be one-fourth as much.

Learn more about gravitational force:

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Setler79 [48]

Answer:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy ( $E$ ) is the sum of gravitational potential energy ( $U$ ) and translational kinetic energy ( $K$ ), all measured in joules. In addition, gravitational potential energy is directly proportional to height ( $h$ ) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

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2 years ago

In the past, most children who went sledding in the winter snow in Verland used wooden sleds with runners and steering bars. Ten

Dovator [93]

Answer:

(E) it seems like the best option

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2 years ago

A computer disk drive is turned on starting from rest and has a constant angular acceleration. If it took 0.750 seconds for the

Irina-Kira [14]

Answer:

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Explanation:

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3 years ago

6. A 50 N block is raised 2 m. If the net work done on the block is 50 J, what is the applied force on the block?

geniusboy [140]

Answer:

F = 75[J]

Explanation:

We know that work is defined as the product of force by distance.

In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

$W = -(F_{weight*d})+(F_{upward}*d)$

where:

W = work = 50 [J]

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Fupward [N]

Now replacing:

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3 years ago

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alina1380 [7]

Answer:

Explanation:

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3 years ago

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