Ask question

Ask question

All categories

Sergio039 [100]

3 years ago

8

A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th

e capacitor is charged by a battery to 575 V and then is disconnected from the battery.
(a) How much energy is stored in the capacitor?
____________J
(b) The separation between the plates is now increased to 10.04 mm. How much energy is stored in the capacitor now?
____________J
(c) How much work is required to increase the separation of the plates from 2.51 mm to 10.04 mm?
_____________ J

2 answers:

anastassius [24]3 years ago

8 0

Answer:

(a) Energy stored is 2.63×10^-5 J

(b) Energy stored is 6.56×10^-6 J

(c) Work required is 8.75×10^-6 J

Explanation:

(a) C = AEo/d

A is area of plates = 451 cm^2 = 451/10000 = 0.0451 m^2

Eo is permutivity constant = 8.84×10^-12 F/m

d is separation between the plates = 2.51 mm = 2.51/1000 = 2.51×10^-3 m

C = 0.0451 × 8.84×10^-12/2.51×10^-3 = 1.59×10^-10 F

E = 1/2CV^2 = 1/2 × 1.59×10^-10 × 575^2 = 2.63×10^-5 J

(b) d = 10.04 mm = 10.04/1000 = 0.01004 m

C = 0.0451 × 8.84×10^-12/0.01004 = 3.97×10^-11 F

E = 1/2CV^2 = 1/2 × 3.97×10^-11 × 575^2 = 6.56×10^-6 J

(c) d = 10.04 mm - 2.51 mm = 7.53 mm = 7.53/1000 = 7.53×10^-3 m

C = 0.0451 × 8.84×10^-12/7.53×10^-3 = 5.29×10^-11 F

W = 1/2CV^2 = 1/2 × 5.29×10^-11 × 575^2 = 8.75×10^-6 J

denpristay [2]3 years ago

3 0

Answer:

A. E = 2.63×10^-5 J.

B. E = 6.56×10^-6 J.

C. W = 8.75×10^-6 J.

Explanation:

A.

C = A * Eo/d

Where,

Eo = free space permutivity

= 8.84×10^-12 F/m

d = distance between the plates

A = area of plates

= 451 cm^2.

Converting from cm to m,

100 cm = 1 m

= 451 * 1/(100)^2 cm

= 0.0451 m^2

d = 2.51 mm

Converting from mm to m,

1000 mm to m

= 2.51 mm * 1 m/1000 mm

= 2.51 × 10^-3 m

Therefore,

C = 0.0451 × (8.84 × 10^-12/2.51 × 10^-3)

= 1.59 × 10^-10 F

E = 1/2 * C * V^2

= 1/2 × 1.59 × 10^-10 × (575^2)

= 2.63 × 10^-5 J.

B.

d = 10.04 mm

Converting from mm to m,

1000 mm = 1 m

= 10.04 * 1 m/1000 mm

= 0.01004 m.

C = 0.0451 × (8.84 × 10^-12/0.01004)

= 3.97 × 10^-11 F

E = 1/2CV^2

= 1/2 × 3.97 × 10^-11 × (575^2)

= 6.56 × 10^-6 J

C.

d = 10.04 mm - 2.51 mm

= 7.53 mm

Converting from mm to m,

1000 mm = 1 m

= 7.53 * 1 m/1000 mm

= 7.53 × 10^-3 m

C = 0.0451 × (8.84 × 10^-12/7.53×10^-3)

= 5.29 × 10^-11 F

W = 1/2 * C * V^2

= 1/2 × (5.29 × 10^-11 × (575^2))

= 8.75 × 10^-6 J

You might be interested in

Describe how acid rain<br> would react with rocks<br> over time.

Papessa [141]

Answer:

They break down rocks.

Explanation:

The rain is slightly acidic due to the presence of carbon dioxide. Power plants and automobiles produce gases such as sulfur dioxide and nitric oxide, which react with water vapor in the atmosphere to form acid rain. They help in weathering the rocks which makes up the Earth's crust.

Hope this helps

:)

3 0

3 years ago

Who watching all star draft? Luka better get picked first ong

Whitepunk [10]

Na dame Dolla fo surrrreeeeee

5 0

3 years ago

Read 2 more answers

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

What is necessary for a substance to be a conductor of electricity?

Stella [2.4K]

The main requirement for a good conductor of electricity is to have a lot of valence electrons. Valence electrons are the electrons of the outer shells of atoms not bound with other atoms (for example through covalent bounds). These electrons are "free to escape" as soon as an electric field with enough intensity is applied to the material, and therefore these electrons will be free to move in the material producing an electric current.

3 0

4 years ago

75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is

mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0

3 years ago