To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

Here,
q = Charge
t = Time
R = Resistance
C = Capacitance
When the charge reach its half value it has passed 10ms, then the equation is,




We know that RC is equal to the time constant, then

Therefore the time constant for the process is about 14ms
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
Answer:
<u>954.4m/s</u>
Explanation:
For a free falling object,it has constant acceleration and a changing velocity.
By using the velocity-time formula, the velocity can be obtained.
The height the rock travelled is the distance.
From,
Velocity (v) = Distance (d) / Time(t)
v = 3245m/3.4s
v = <u>954.4m/s</u>
That js the answer I got. Hope it's right.
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