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
Explanation:
It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.
“A place where things are baked”
- the bakery?
The period of the transverse wave from what we have here is 0.5
<h3>How to find the period of the transverse wave</h3>
The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.
The formula with which to get the period is
w = 4π
where w = 4 x 22/7
2π/T = 4π
6.2857/T = 12.57
From here we would have to cross multiply
6.2857 = 12.57T
divide through by 12.57
6.2857/12.57 = T
0.500 = T
Hence we can conclude that the value of T that can determine the period based on the question is 0.500.
Read more on transverse wave here
brainly.com/question/2516098
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