(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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Work = force x distance.
force = mass x acceleration
work = mass x acceleration x diastance
use acceleration of gravity in this problem
W (J) = m (kg) x a (m/s/s) x d (m)
W = 78 x 9.8 x 6
W = 4586.4
Answer:
Explanation:
The formula for the magnitude of a vector is
and then round to the hundredths place:
3.11 m. Since we are in Q2, we can also find the direction of this vector:
but since we are in Q2, we add 180 degrees to the result, getting the angle to be 115.3
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance

r is the radius of the spherical shell
dr is the thickness
volume of shell

mass of shell


now,

integrating both side



we know,




a = 9.94 m/s²
Answer:
The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.
C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'
Explanation:
See attached picture.