(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
#SPJ1
The ball will take 2.551 seconds to reach its peak position.
<h3>How much time will the ball take to land?</h3>
We must know how long the balls are in the air before we can predict where they will fall. It will take 2 seconds for both balls to touch the ground.
<h3>How quickly does a ball drop?</h3>
The falling ball travels a distance of d = 12 9.8 (m/s2) t2, with a speed of v = 9.8 (m/s2) t as a function of time. The ball travels 4.9 m in a second. The falling ball's velocity is v = -9.8 (m/s2) t j, and its position is r = (4.9 m - 12 9.8 (m/s2) t2) j as a function of time.
To know more about balls visit:-
brainly.com/question/19930452
#SPJ4
It stands for Unified Computing System
I hope I helped :3
Answer:
whats the question... you didn't put the question up
Answer:
The maximum power delivered by the power supply is 0.81 W.
Explanation:
Given that,
Inductance L= 2.0 H
Resistance R = 100 ohm
Voltage = 9.0 V
We need to calculate the power
Using formula of power
Where, P = power
V = voltage
R = resistance
Put the value into the formula
Hence, The maximum power delivered by the power supply is 0.81 W.
