Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
Answer:
False
Explanation:
The formula of force that exists between two charges is expressed as;
F = kq1q2/r²
If two charges separated by one meter exert a 9 N force on each other, the;
9 = kq1q2/1²
9 = kq1q2 ..... 1
If the charges are pushed to a 3 meter separation, then;
F = kq1q2/3²
F = kq1q2/9 .... 2
Divide both equations;
9/F = (kq1q2)/ kq1q2/9
9/F = kq1q2 * 9/ kq1q2
9/F = 9
F = 9/9
F = 1N
Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False
Answer:
kinetic energy=1/2mv^2.
which is 4320000=1/2×m×23^2.
which is 4320000=1/2×m×529.
4320000=264.5m.
m=4320000/264.5.
m=16332.70~16333g
B. I think is the correct answer
