(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:
The energy becomes 4 times greater.
Explanation:
We know that the energy of a wave is proportional to the square of its amplitude
E ∝ Amplitude^2
Since the original amplitude = 0.5 m
and the new amplitude becomes = 1 m
We are doubling the amplitude. This means that the new energy will be affected by a factor of 4
E_new ∝ (2*Amplitude)^2 =
E_new ∝ 4*(Amplitude)^2
E_new = 4*E
Answer: I = 3.6 m3
(C)
Explanation:
moment of inertia for spherically shaped object around it's center is given as
I = (2/5) mr²
substituting the r = 3m²
I = (2/5)*(9) m3
I = 3.6 m3
Answer:
Explanation:
The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.
Let us find the acceleration:
Electric force is given as the product of charge and electric field strength:
F = qE
where q = electric charge
E = Electric field strength
Force is generally given as:
F = ma
where m = mass
a = acceleration
Equating both:
ma = qE
E = ma / q
For an electron:
m = 9.11 × 10^{-31} kg
q = 1.602 × 10^{-19} C
Therefore, the electric field strength of the electron is:
