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exis [7]
4 years ago
10

How much charge flows from a 6.0 v battery when it is connected across a completely discharged 15.7 μf capacitor?

Physics
1 answer:
Lilit [14]4 years ago
4 0
The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 6.0V (assuming circuit resistance is negligable) and it has a capacitance of 15.7<span>μf or </span> 15.7x10^-6f, therefore charge equals (15.7x10^-6)x6=9.42x10^-5C (Coulombs).


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A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
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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

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

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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  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

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

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

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

v = √8.1534

v = 2.855 m/s

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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2 years ago
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AnnyKZ [126]
The statement that can be used to answer this  question is:

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