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Answer :
(a) The magnitude of the net gravitational force exerted by the 195 kg and 495 kg objects on the 65.0 kg object is approximately 1.49 x 10⁻⁷ N.
(b) The 65.0 kg object can be placed at a position approximately 1.80 m from the 495 kg object toward the 195 kg object to experience a net force of zero from the other two objects.
(a) The gravitational force between two objects can be calculated using the equation:
F = G * (m₁ * m₂) / r²
where F is the gravitational force, G is the gravitational constant (approximately 6.67 x 10⁻¹¹N*m²/kg²), m₁ and m₂ are the masses of the objects, and r is the distance between their centers of mass.
In this case, the net gravitational force on the 65.0 kg object is the sum of the gravitational forces exerted by the 195 kg and 495 kg objects. Since the objects are placed midway between them, the distance between the 65.0 kg object and each of the other objects is 4.70 m / 2 = 2.35 m.
Using the equation for gravitational force, we can calculate the force exerted by each object on the 65.0 kg object, and then find the net force by taking the sum:
F_net = F_195 + F_495
F_net = G * ((m₁ * m₃) / r₁² + (m₂ * m₃) / r₂²)
Plugging in the given values, we find:
F_net ≈ 1.49 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by the 195 kg and 495 kg objects on the 65.0 kg object is approximately 1.49 x 10⁻⁷ N.
(b) To find the position where the net force on the 65.0 kg object is zero, we need to set up an equation using the gravitational forces exerted by the 195 kg and 495 kg objects.
The gravitational force exerted by the 195 kg object on the 65.0 kg object can be calculated using the equation:
F_195 = G * ((m₁ * m₃) / r₁²)
Similarly, the gravitational force exerted by the 495 kg object on the 65.0 kg object is given by:
F_495 = G * ((m₂ * m₃) / r₂²)
For the net force to be zero, the magnitudes of these two forces must be equal:
F_195 = F_495
G * ((m₁ * m₃) / r₁²) = G * ((m₂ * m₃) / r₂²)
Simplifying the equation and substituting the given values, we can solve for the position:
r₂² = (m2 * r₁²) / m₁
r₂ ≈ √((m * r₁²) / m₁)
r₂ ≈ √((495 kg * (2.35 m)²) / 195 kg)
r₂ ≈ 1.80 m
Therefore, the 65.0 kg object can be placed at a position approximately 1.80 m from the 495 kg object toward the 195 kg object to experience a net force of zero from the other two objects.
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The complete question is:
A 195-kg object and a 495-kg object are separated by 4.70 m.
(a) Find the magnitude of the net gravitational force exerted by these objects on a 65.0-kg object placed midway between them. N (
b) At what position (other than an infinitely remote one) can the 65.0-kg object be placed so as to experience a net force of zero from the other two objects? m from the 495 kg mass toward the 195 kg mass
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