If the distance between two charge particles is increased by four, the force will decrease by a factor of?

According to Coulomb's law, the magnitude of electrostatic force of attraction between two point charges is directly proportional to product of magnitudes of charges and inversely proportional to the square of the distance between them.

Mathematically,      F = \frac{q_{1}q_{2}}{r^{2}}

where         F = force

                q_{1} = charge 1

                q_{2} = charge 2

               r = distance between the charges

Therefore, when distance is increase by four times then the force will decrease by a factor of 16 as force is inversely proportional to the square of distance.

Explanation:

Hope this helps :)

the titanic would be the correct answer

answer-

the statement that f(x) = |x+a| + b has exactly one x-intercept is sometimes correct.

solution-

it solely depends on b, whether the function will have one or two or zero x intercept.

this plot of the given function, f(x) = |x+a| + b will be the basic absolute value graph i.e v shape, with vertex translated to (-a, b), instead of origin.

1-   if b is zero, the graph will have exactly one x-intercept, at x= -a

2- if b is positive, the whole graph will be above the x-axis, hence it will have no x-intercepts.

3- if b is negative, the graph will be below the x-axis, hence it will have two x-intercepts.

∴ the statement is sometimes true.