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
When driver see the child standing on road his speed is 20 m/s
So here at that instant his reaction time is 0.80 s
He will cover a total distance given by product of speed and time



now after this he will apply brakes with acceleration a = 7 m/s^2
so the distance covered before it stop is given by



so the total distance covered by it


<em>so it will cover a total distance of 44.6 m</em>
Answer:

Explanation:
The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

We have
. Thus, solving for t:

The answer is c 1386j
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