Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
<h2>
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>
Explanation:
One period means time taken to complete one revolution.
In case of swings in one period time it travels the same position through two times.
Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.
For 1 period = 2 times
For 3 periods = 3 x For 1 period
For 3 periods = 3 x 2 times
For 3 periods = 6 times
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.
Answer:
B.The box experiences less friction on the marble floor
Explanation:
