Two charged ping-pong balls separated by a distance of 1.63 m exert an electric force of 0.0224 n on each other. what will be the force if the objects are brought closer, to a separation of only 24.0 cm? (in n)

1.024 N

Explanation:

Step 1:

Using Coulomb's law: F=kQ1Q2/d^2

We can rearrange the terms to get :

kQ1Q2=Fxr^2

Substituting: kQ1Q2=0.0224N x (1.63m)^2

Thus kQ1Q2=0.0595

We know that this value will not change when the balls are brought closer together.

Step 2:

Change from cm to m : 1cm=0.01m

24cm=0.24 m

Substitute into the new distance and the calculated value for kQ1Q2 into the coulomb law formula to calculate the new force

F=kQ1Q2/d^2=0.059/(0.24)^2

F=1.024N

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas