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Maksim231197 [3]
3 years ago
7

Gravitational Force: Two small objects, with masses m and M, are originally a distance r apart, and the magnitude of the gravita

tional force on each one is F. The masses are changed to 2m and 2M, and the distance is changed to 4r. What is the magnitude of the new gravitational force?
Physics
1 answer:
Alona [7]3 years ago
4 0

Law of universal gravitation:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m = masses of the objects, r = distance between the objects

F is proportional to both M and m:

F ∝ M, F ∝ m

F is proportional to the inverse square of r:

F ∝ 1/r²

Calculate the scaling factor of F due to the change in M:

k₁ = 2M/M = 2

Calculate the scaling factor of F due to the change in m:

k₂ = 2m/m = 2

Calculate the scaling factor of F due to the change in r:

k₃ = 1/(4r/r)² = 1/16

Multiply the original force F by the scaling factors to obtain the new force:

Fk₁k₂k₃

= F(2)(2)(1/16)

= F/4

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Explanation:

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          v_{3y} = \frac{21}{2}

                     = 10.5 m/s

2mv_{3x} = m(21)

          v_{3x} = \frac{21}{2}

                     = 10.5 m/s

  v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}

              = \sqrt{(10.5)^{2} + (10.5)^{2}}

              = 14.84 m/s

or,          = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

          \theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})

                      = tan^{-1}(\frac{10.5}{10.5})

                      = tan^{-1} (45^{o})

                      = 45^{o}

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Answer:

a = 0.53 m/s^2

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

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\omega = 2\pi ( \frac{20}{60})

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so final tangential speed is given as

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v = 1.71 (2.10) = 3.58 m/s

now average acceleration of the girl is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{3.58 - 0}{6.73}

a = 0.53 m/s^2

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8/4 = y/y-x

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Hope this helps
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