Force f acts between a pair of charges, q1 and q2, separated by a distance d. for each of the statements, use the drop-down menus to express the new force in terms of f.

q1 is halved, q2 is doubled, but the distance between the charges remains d.
f/4
f
6f
6f/4

q1 and q2 are unchanged. the distance between the charges is doubled to 2d.
f/4
f
6f
6f/4

q1 is doubled and q2 is tripled. the distance between the charges remains d.
f/4
f
6f
6f/4

The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

So, the new force is:

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

So, the new force is:

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

So, the new force is:

So the force has increased by a factor 6.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer