Three masses that are 10 kg each are located at the corners of an equilateral triangle,with side length of 0.55 m.What is the magnitude of the total force acting on one mass due to the other two masses?

Question

Quizplus · Accepted Answer

The correct answer is A because the magnitude of the total force acting on one mass due to the other two masses is 3.82 * 10^-8 N. This is because the force between two masses is given by the equation F = Gm1m2/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. In this case, the masses are each 10 kg, the distance between them is 0.55 m, and the gravitational constant is 6.674 * 10^-11 N m^2 / kg^2. Plugging these values into the equation, we get F = (6.674 * 10^-11 N m^2 / kg^2) * (10 kg) * (10 kg) / (0.55 m)^2 = 3.82 * 10^-8 N.

Three Masses That Are 10 Kg Each Are Located at the Corners