1.)two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. what is the force of gravitation between these two objects? 2.) if the two objects in the previous question are moved so that the distance between them is cut exactly in half to 2.5 m, what is the new value of the force between them? 3.) if a perpendicular force is applied to an object at 0.5 m from the radius, causing it to rotate, what is the torque if the force is 25 n? 4.)a hoop that has a moment of inertia (i) of 0.007875 kg*m^2 and an angular velocity of 6.28 rad/s has what value for angular momentum (l)?

1)

The force of gravitation between the two objects is given by:

where

is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

2)  

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

So, the gravitational force increases by a factor 4. Therefore, the new force will be

3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity () is:

In this problem, we have

So, the angular momentum is