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a 175-kg object and a 475-kg object are separated by 3.90 m. (a) find the magnitude of the net gravitational force exerted by these objects on a 53.0-kg object placed midway between them. n (b) at what position (other than an infinitely remote one) can the 53.0-kg object be placed so as to experience a net force of zero from the other two objects? m from the 475 kg mass toward the 175 kg mass

Sagot :

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a) The magnitude of the net gravitational force exerted by these objects on a 53.0 kg is 6.03 * 10¹¹ N

b) By placing object of 53.0 kg 1.976 m from 175 kg object and 1.324 m from 475 kg object.

Mass of an object-1= M₁ = 175 kg

Mass of an object-2= M₂ = 475 Kg

Distance between M₁ & M₂  = 3.90 m

Mass of an object-3 = M₃ = 53 Kg

If M₃ placed midway between them M₁ & M₂.

Force on M₃ due to M₁

Distance between  M₁ & M₃, r = 3.90/2 = 1.95

F = G(M₁*M₃/r²)

F = 6.67 x 10¹¹(175*53/ 1.95²)

F = 16,269.33*10¹¹ = 1.62 * 10⁷ N

Force on M₃ due to M₂

Distance between M₂ & M₃, r' = 3.90/2 = 1.95

F' = G(M₂*M₃/r²)

F = 6.67 * 10₁₁ ( 475 * 53/ 3.8)

F = 44188.75 * 10¹¹ = 4.41 * 10⁷ N

Net force on M₃:

F' + F = 4.41 * 10⁷ N + 1.62 * 10⁷ N = 6.03 * 10¹¹ N

The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is 6.03 * 10¹¹ N.

b) Distance between M₁ & M₂  = r = 3.90 m

Let the distanced between M₁ & M₃ be r

Force experienced by M₃ due to M₁ :

f = G(M₁*M₃/r²)

Let the distanced between M₂ & M₃  be r'

Force experienced by M₃ due to M₂:

f' = G(M₂*M₃/r²)

Net force equal to zero, then :

f = f'

G(M₁*M₃/r²) = G(M₂*M₃/r'²)

6.18*10⁷/r₂² = 16.7*10₇/r'²

r = 1.4914 * r'

R = r+r'

3.90 m = 1.4914\times r' + r'

r' = 1.324 m

r = 1.976 m

Therefore, the magnitude of the force on M₃ is 6.03 * 10¹¹ N and by placing object of 53.0 kg 1.976 m from 175 kg object and 1.324 m from 475 kg object.

To know more about force, refer: https://brainly.com/question/14019879

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