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2 years ago

How does the gravitational force between two objects change if the distance between the objects doubles?

Physics

1 answer:

Ymorist [56]2 years ago

4 0

if the distance between the objects is doubled the force is reduced by a factor of 4

<h3>Whats is gravitational force?</h3>

Gravitational force is the force of attraction between objects in the universe.

f = G * m1 * m2 / r^2

f = gravitational force

G = gravitational constant

m1 = mass of object 1

m2 = mass of object 2

r = distance between the objects

From the formular, the gravitational force and the distance is an inverse relationship so increasing the distance by a factor results to reduction of the force by the square of the factor. hence doubling the distance which is distance mutiplied by 2 will lead to reduction of the force by 2^2 = 4

Therefore: The force decreases by a factor of 4.

hope it helps

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Read 2 more answers

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

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(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

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The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision

In this situation, we have

$\Delta t = 0.06 s$

So we can re-arrange the equation to find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N$

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3 years ago