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Calculating the Force of Attraction Between Rocks at a Reduced Distance

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Question: Two rocks are close to each other in space. An attractive force of 4 newtons (N) acts on each of the rocks. If the distance between the rocks is reduced to 250 meters (m), what would be the force of attraction acting on each rock?

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The force of gravity between two objects in space is given by Newton's Law of Universal Gravitation: F=G*(m1*m2)/d^2 where: - F is the force between the masses, - G is the gravitational constant, - m1 and m2 are the masses of the two objects, and - d is the distance between the centers of the two objects. From your question, we can see that the initial force F is 4 N and the initial distance d is unknown. When the distance is reduced to 250 m, the new force F' and the new distance d' are to be determined. Since we don't know the masses m1 and m2, we can't calculate the exact force. But we can say that the force is proportional to the inverse square of the distance. F/d^2 = F'/d'^2 We know that F = 4 N and d' = 250 m, but without knowing the initial distance or the masses of the objects, we can't say exactly what F' would be. However, if the distance is reduced, the force will increase, because they are inversely proportional. They behave in such a way that if one is halved, the other is quadrupled. If you could provide more information, I would be able to give a more precise answer.

March 22, 2024, 10:19 a.m.

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