High School

Objects with masses of 181 kg and 712 kg are separated by 0.442 m. A 72.6 kg mass is placed midway between them.

Find the magnitude of the net gravitational force exerted by the two larger masses on the 72.6 kg mass.

The value of the universal gravitational constant is \(6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\).

Answer in units of N.

Answer :

Final answer:

The magnitude of the net gravitational force exerted by the two larger masses on the 72.6 kg mass is 5.978 N.

Explanation:

The magnitude of the net gravitational force exerted by the two larger masses on the 72.6 kg mass can be calculated using Newton's Law of Universal Gravitation. The formula is given as F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the masses.

Plugging in the values: G = 6.672 x 10^-11 N · m^2/kg^2; m1 = 181 kg; m2 = 712 kg; r = 0.442 m

We can now calculate the gravitational force using the formula: F = (6.672 x 10^-11 N · m^2/kg^2) * (181 kg * 72.6 kg) / (0.442 m)^2

Simplifying the calculation, we get F = 5.978 N (rounded to three decimal places).

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