(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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Just do energy spent divided by time to get your answer. With this we can say a human might be able to!
Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s
Explanation:
For proton, the centripetal force required for circular motion is provided by the magnetic force,
so Fm= Fc
q v B = m v²/r
m= mass of charged particle
v= velocity
B =magnetic field
q= charge
r= radius of circular path
v= q B r/m
now v= r ω
ω= angular velocity
ω r = q B r /m
ω=q B /m
now ω= 2π/T where T =time period
so 2π/T=q B/m
T= 2 πm/q B
T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]
T= 7.28 x 10⁻⁸ s
Navigators used astrolabes to calculate positions of the sun and stars.
