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
Answer:
When it comes to serving, the court is divided into six zones. Right back is zone one, right front is zone two, middle front is zone three, left front is zone four, left back is zone five and middle back is zone six.
Explanation:
Answer: see the graph attached (straight line, passing through the origin and positive slope).
Justification:1)
Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x:
y α xii) That implies:
y = kx,where k is the constant of proportionality.
iii) The graph is a
line that passes through the origin and has positive slope k (k = y / x).2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the
Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.
250kg
would have momentum that is being caried by the impact of the trow