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Georgia [21]
3 years ago
8

A capacitor is discharged through a 30.0 Ω resistor. The discharge current decreases to 26.0% of its initial value in 3.00 ms .

What is the value of the capacitor?
Physics
1 answer:
mestny [16]3 years ago
5 0

Answer:

74.23\mu F

Explanation:

We have given the value of resistance R = 30 ohm

Time in which current will decreases to 26% is 3 ms

The discharging equation of the capacitor is given by

i=i_0e^\frac{-t}{\tau } here \tau is time constant which is given as \tau =RC

So 0.26i_0=i_0e^\frac{-3}{\tau }  

0.26=e^\frac{-3}{\tau }

\frac{-3}{\tau }=ln0.26

\frac{-3}{\tau }=-1.347

{\tau }=\frac{3}{1.347}=2.2271ms

WE know that \tau =RC=30C

C=\frac{2.227}{30}=0.074mF=74.23\mu F

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

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The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

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rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

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a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

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α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

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ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

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