Two particles experience an attractive electric force of 5 N. If one of the charges triples and the distance between them doubles, what is the new electric force?

3.75 N

Explanation:

F = k(q)(q)/r^2 = 5 N

F' = k(q)(3q)/(2r)^2

= k×3(q)(q)/4r^2

= (3/4)[k(q)(q)/r^2]

= (3/4)F

= (3/4)(5 N)

= 3.75 N

electromagnetic waves travel through space with vacuums. give me brainliest if right.

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