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Leaving the distance between the 99 kg and the 404 kg masses fixed, at what distance from the 404 kg mass (other than infinitely remote ones) does the 27.9 kg mass experience a net force of zero?

Answer in units of meters.

Answer :

The 27.9 kg mass experience a net force of zero. Gravitational equilibrium occurs when the gravitational force on an object is equal to zero.

How is equilibrium impacted by gravity?

Tilting an object will increase the height of its center of gravity when it is in a stable balance. When tilted, an item in unstable equilibrium will have a lower center of gravity. Finally, for objects in neutral equilibrium, the center of gravity will stay at the same height when pushed.

How can we determined this?

To find the distance at which the 27.9 kg mass experiences a net force of zero, we need to calculate the gravitational force acting on it due to the other two masses. The gravitational force between two masses m1 and m2 at distance r is given by F = Gm1m2/r^2, where G is the gravitational constant.

First, we know that the distance between the 99 kg and 404 kg masses is fixed. Let's call this distance d.

Now we can set the gravitational force acting on the 27.9 kg mass due to the 99 kg mass and the 404 kg mass equal to zero and solve for r:

0 = G*(27.9 kg)(99 kg)/r^2 + G(27.9 kg)*(404 kg)/(d-r)^2

Solving for r, we get:

r = d * sqrt(404/(404+99))

Substituting d value will give the distance at which the 27.9 kg mass experiences a net force of zero in units of m.

To know more about gravitational force visit:

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