A 10.0 μF capacitor initially charged to 20.0 μC is discharged through a 2.80 kΩ resistor. How long does it take to reduce the capacitor's charge to 10.0 μC ?

19.4 ms

Step-by-step explanation:

The electric charge in a capacitor is given as;

Q= CV

Initial volts V = Q/C = 20µC/10µF = 2volts

final volts V = Q/C= 10µC/10µF = 1volts

Voltage on a capacitor undergoing discharge is given as;

v = v₀e^(–t/τ)

Where;

v₀ = initial voltage on the capacitor

v = voltage after time t

R = resistance in ohms,

C = capacitance in farads

t = time in seconds

RC = τ = time constant

Therefore;

τ = 10µF x 2.8kΩ = 28 ms

v = v₀e^(–t/τ)

1 = 2e^(–t/28ms)

e^(–t/28ms) = 0.5

–t/28ms = -0.693

t = (28ms)(0.693) = 19.4 ms

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.