(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
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Answer:
C
Explanation:
Because...
A= Incideant ray
B= Angle of inciderance
<u>C= angle of reflection</u>
D= reflection ray
Explanation:
the speed of the motor can be increased by increasing the strength magnetic field increasing the current flowing through the coil.
Answer:
the spring constant is -89.2857 n/m
Explanation:
The computation of the spring constant is shown below:
As we know that
Force in newtons = Spring constant × amount of extension
F = -k × x
where
F = 25 N
And, x = 0.280m
So, the spring constant would 2
= 25N ÷ 0.280 m
= -89.2857 n/m
Hence the spring constant is -89.2857 n/m
Answer:
the law of conservation of energy
