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sergiy2304 [10]
3 years ago
13

What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not

change(b) It is multiplied by 2(c) It is multiplied by 4(d) It is multiplied by 8What happens to the gravitational force between two particles if the distance between them is doubled? (a) It does not change(b) It decreases by a factor of 2(c) It decreases by a factor of 4(d) It decreases by a factor of 8
Physics
1 answer:
mr_godi [17]3 years ago
7 0

Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

F=G*\frac{m1*m2}{r^{2} }

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives  the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }

F'=G*\frac{m1*m2}{\frac{(r )^{2} }{2^{2} } }

F'=G*\frac{m1*m2}{\frac{(r )^{2} }{4} }

F'=4*G*\frac{m1*m2}{r^{2} }

F'=4*F

<u><em> The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).</em></u>

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Answer:

magnitude of the force exerted on the wall is 2100 N.

Explanation:

given data

rate = 50.0 kg/s

speed = 42.0 m/s

solution

momentum is reduced = zero

so final speed v = 0

so relation between the force and the momentum is express as

Force =  \dfrac{p}{t}    ...............1

Force = \dfrac{mv}{t}    

put here value and we get

Force = 50 \times 42  

Force = 2100 N

so that magnitude of the force exerted on the wall is 2100 N.

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What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes
bagirrra123 [75]

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

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  • r is distance of separation
  • k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

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F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

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HELP!
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Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

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It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

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Answer:

\boxed {\tt 3.6 \ seconds}

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t=\frac{45 \ m}{12.5 \ m/s}

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