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dybincka [34]
3 years ago
9

According to Newton’s law of universal gravitation, as the mass or one or both bodies increases, what happens to the force of at

traction between the bodies?
force decreases

causes no change

force increases
Physics
1 answer:
wel3 years ago
3 0

Answer:

force increases

Explanation:

Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.

According to Newton’s law of universal gravitation, as the mass or one or both bodies increases, the force of attraction between the bodies increases.

Mathematically, this is given by the Newton's Second Law of Motion;

Force = mass * acceleration.

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Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
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Given:

- Attachment for figures missing in the question.

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                                         E = k*Q / r^2

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Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

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                                  E_net = 2*E_part(a)*cos(Q)

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                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

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