Apair of electrically charged objects attract each other with a force of 20 n when they are a distance of 5 m apart. if their charges stay the same, then what will be the attractive force between them when they are 20 m apart?

Assume that the two charges are q1 and q1.
The force of attraction between the two charges is given by the following equation:
F = (k * q1 * q2) / (r^2)  
F1 = 20n
r1 = 5m and r2 = 20m
We can notice that r2 = 4r1
therefore,
F2 = (k * q1 * q2) / [(4r)^2]
F2 = (1/16) x F1
Since F1 = 20n, therefore,
F2 = (1/16) x 20 = 1.25n 

Yes

He can play two games. 13/4=3.253.25*2=6.5