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NemiM [27]
3 years ago
11

What happens to the force of gravitational pull if we tripled the distance and one of the masses is doubled

Physics
2 answers:
Svetllana [295]3 years ago
5 0
Triple the distance ... The force drops to 1/9 of the original force.

Double one mass ... The original force doubles.

Do both ... The original force becomes 2/9 as strong, = about 22.2% of what it was originally.
kicyunya [14]3 years ago
4 0
This depends on the original mass of the object having its mass doubled and the the original distance before the distance was tripled.
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Answer:

When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.

Explanation:

Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.

Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.

When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.

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Which of these statements explains the difference between nuclear binding energy and the strong nuclear force? Check all that ap
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3 years ago
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ollegr [7]

Answer:

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Explanation:

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A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he
RideAnS [48]

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

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Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

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we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

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3 years ago
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IgorLugansk [536]
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2 years ago
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