3 years ago

What happens to the gravitational attraction between 2 objects

Physics

1 answer:

amm18123 years ago

7 0

Answer:

It decreases.

Explanation:

between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases

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If the time of impact in a collision is extended by 3 times, by hour much is

vodomira [7]

Answer:

The force of the impact would be smaller

Explanation: Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

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3 years ago

I just need the answer

Archy [21]

Answer:

I'll try to help which grade are you?

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3 years ago

lonnie pitches a baseball of mass 0.02kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to L

vodomira [7]

Answer:

I=2 kg.m/s

Explanation:

The impulse is defined as the change of momentum:

$I=p_f-p_o\\I=m*v_f-m*v_o\\I=0.02kg*[(-60m/s)-40m/s]\\I=2kg.m/s$

We took the final velocity as negative since it is going on the opposite direction of the intial motion of the ball.

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3 years ago

A ball is thrown upward from the ground with an initial speed of 22.0 m/s; at the same instant, another ball is dropped from a b

Vikki [24]

They will be together in 8 minutes

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3 years ago

A small artery has a length of 1.10 × 10-3 m and a radius of 2.50 × 10-5 m. If the pressure drop across the artery is 1.15 kPa,

levacccp [35]

Answer:

$7.69533\times 10^{-11}\ m^3/s$

Explanation:

P = Pressure difference = 1.15 kPa

r = Radius = $2.5\times 10^{-5}\ m$

$\eta$ = Viscosity of liquid = $2.084\times 10^{-3}\ Pas$

l = Length of artery = $1.1\times 10^{-3}\ m$

From Poiseuille's equation we have

$Q=\frac{\pi Pr^4}{8\eta l}\\\Rightarrow Q=\frac{\pi 1.15\times 10^3\times (2.5\times 10^{-5})^4}{8\times 2.084\times 10^{-3}\times 1.1\times 10^{-3}}\\\Rightarrow Q=7.69533\times 10^{-11}\ m^3/s$

The flow rate of blood is $7.69533\times 10^{-11}\ m^3/s$

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3 years ago