If the distance between a pair of electric charges is cut in half, the electrostatic force changes in size by : a factor of four
a factor of one-quarter
a factor of two
a factor of one-half

If the distance between a pair of electric charges is cut in a half, the electrostatic force changes in size by a factor of four.

Explanation:

The electrostatic force between the charges can be calculated with the Coulomb's law, that in its scalar form is given by:

Where   is the Coulomb's constant; and are the electric charges; and is the distance between the charges. From the equation above, you can establish that the force and the distance between the charges have are inversely proportional.

If the distance between the charges is cut in half, then you have:

Organizing and doing small calculations:

So, you can see that the resultant force changes in size by a factor of four.

resonant frequency

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft