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Anna007 [38]
3 years ago
8

A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by th

ese objects on a 42.0-kg object placed midway between them.
Physics
1 answer:
Eddi Din [679]3 years ago
7 0

Answer:

The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

Explanation:

Given;

first object with mass, m₁ = 285 kg

second object with mass, m₂ = 585 kg

distance between the two objects, r = 4.3 m

The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m

Gravitational force between the first object and the 42 kg object;

F = \frac{GMm}{r^2}

where;

G = 6.67 x 10⁻¹¹ Nm²kg⁻²

F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N

Gravitational force between the second object and the 42 kg object

F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N

Magnitude of net gravitational force exerted on 42kg object;

F = 3.545x 10⁻⁷ N  -  1.727 x 10⁻⁷ N

F = 1.818 x 10⁻⁷ N

Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

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