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PSYCHO15rus [73]
3 years ago
9

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

.0-Ω resistor is connected between the two terminals of the capacitor. Find the charge on the capacitor at the following times. (When entering units, use micro for the metric system prefix µ.)
Physics
1 answer:
Grace [21]3 years ago
4 0

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

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a hot iron bar is placed 100ml 22C water. the water temperature rises to 32C. how much heat did the water gain and how much heat
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Answer:

  • Water gained: 10
  • Iron lost: -10

Explanation:

Given: Hot iron bar is placed 100ml 22C water, the water temperature rises to 32C

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Formula:Q = change T x C x M

Solve:

<u>How much heat water gained</u>

Initial heat = 22, then rose to 32. To find how much heat the water gained, simply subtract the current heat by the initial heat.

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<u>How much heat Iron lost</u>

Current heat = 32, then dropped to 22. To find how much heat the Iron lost, simply subtract the initial heat by the current heat.

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A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
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Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

E_r = \frac{1}{2} mv^2

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

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