Can someone help me plz

A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to

liberstina [14]

(a) $Q = 1.70\times 10^{-2}\;\text{C}$ ;
(b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$ ;
(c) $E_\text{final} = 4.52\;\text{J}$ ;
(d) $\Delta E = 2.82\;\text{J}$ .

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$ .

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$ .

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$ .

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$ .

(d)

Initial energy of the system, which is the same as the initial energy in the $20.0\;\mu\text{F}$ capacitor:

$\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}$ .

Change in energy:

$\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}$ .

4 0

3 years ago

What happens to electrical energy that is used by objects in our homes? (1 point) a It is absorbed by batteries. b It is destroy

Reil [10]

Question:

What happens to electrical energy that is used by objects in our homes? (1 point)

a It is absorbed by batteries.

b It is destroyed.

c It is stored in solar panels.

d It is transformed into other forms of energy.

Answer:

D

3 0

2 years ago

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If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb

Anna [14]

Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

P1V1 = P2V2 ( at temperature is constant )

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg(torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987*(X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L

5 0

3 years ago

What are the two forces that are acting on the skydiver? What is

Cloud [144]

Answer:

Gravity provides a downward force, resulting in the diver going downward. They speed up like any falling object would, the pull of gravity is a dominant force. (There is a drag force – as a result of moving through the air.)

7 0

3 years ago

The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie

Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I' and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0

2 years ago

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