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jekas [21]
1 year ago
6

A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci

tor take to loose half its charge?
B) How long does it take the capacitor to loose half of its stored energy?
Physics
1 answer:
larisa [96]1 year ago
6 0

(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

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EleoNora [17]

Answer: 1.04N

Explanation:

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F =(kq1q2) / r²

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Read 2 more answers
Please answer any of these thanks !
KIM [24]
1).  The equation is: (speed) = (frequency) x (wavelength)

Speed = (256 Hz) x (1.3 m) = 332.8 meters per second

 2).  If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
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If the instrument is played at a higher pitch, then the waves become shorter,
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If the instrument plays louder and at higher pitch, the waves on the scope
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(Notice that this is exactly the same as the equation up above in question #1,
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Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.

That's ' 200 k Hz ' .

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3 years ago
In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every
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Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

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Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

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3 years ago
a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

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W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

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