B. The number of field lines on the source charge.
Answer:
It takes approximately 0.009 seconds for the capacitor to discharge to half its original charge.
Explanation:
Recall that the capacitor discharges with an exponential decay associated with the time constant for the circuit () which in our case is;
(13 milliseconds)
Recall as well the expression for the charge in the capacitor (from it initial value , as it discharges via a resistor R:
So knowing that the capacitor started with a charge of 30 and reduced after a time "t" to 30 , and knowing from our first formula what the RC of the circuit is, we can solve for the time elapsed using the charge as function of time equation shown above:
In seconds this is approximately 0.009 seconds
Answer:
d = 329.81m
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
V_f = (0m/s)+(9.81m/s²)*(8.2s)
V_f = 80.442m/s
d = ((V_f-V_0)/2)*t
d = distance
d = ((80.442m/s-0m/s)/2)*(8.2s)
d = 329.81m
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