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solmaris [256]
3 years ago
10

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu

m foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22×28cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of poster board, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the poster board to use as a dielectric. Will she need a larger or smaller area of Teflon than of poster board? Explain.
Engineering
1 answer:
bazaltina [42]3 years ago
7 0

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

b.

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

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grigory [225]

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

F_{H} = F cos0

F_{H} = (30) x 4/5

F_{H} = 24N

Now Calculate V using following formula

V = V_{0} + at

V_{0} = 0

a = F_{H} / m

a = 24N / 20 kg

a = 1.2m / S^{2}

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = F_{H} x V

Power = 24 x 4.8

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3 0
3 years ago
A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat
jeka94

Answer:

the maximum thermal efficiency is 29%

Explanation:

the maximum efficiency for a thermal engine that works between a cold source and a hot source is the one of a Carnot engine. Its efficiency is given by

Maximum efficiency= 1 - T2/T1

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T2= absolute temperature of the cold sink (environment)= 20°C + 273 = 293

T2= absolute temperature of the hot source (hot water supply) = 140°C + 273 = 413

therefore

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3 0
2 years ago
A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co
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Answer:

a) T_{H} = 1.967\,^{\circ}F, b) COP_{R} = 9.105, c) T_{H} = 115.934\,^{\circ}F, d) COP_{R} = 6.995, e) T_{H} = 25.129\,^{\circ}C

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}

10 = \frac{419.67\,R}{T_{H}-419.67\,R}

The temperature of the hot reservoir is:

10\cdot T_{H} - 4196.7 = 419.67

T_{H} = 461.637\,R

T_{H} = 1.967\,^{\circ}F

b) The coefficient of performance is:

COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}

COP_{R} = 9.105

c) The temperature of the hot reservoir can be determined with the help of the following relation:

\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}

\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}

5 = \frac{479.67\,R}{T_{H}-479.67\,R}

5\cdot T_{H} - 2398.35 = 479.67

T_{H} = 575.604\,R

T_{H} = 115.934\,^{\circ}F

d) The coefficient of performance is:

COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}

COP_{R} = 6.995

e) The temperature of the cold reservoir is:

8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}

8.9\cdot T_{H} - 2386.535 = 268.15

T_{H} = 298.279\,K

T_{H} = 25.129\,^{\circ}C

8 0
2 years ago
What is the mass of a brass axle that has a volume of 0.318 cm? ​
NeX [460]

Answer:

2.7g

Explanation:

the mass of a brass axle that has a volume of 0.318 cm is 2.7g.

8 0
2 years ago
A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
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Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

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The height of pouring cavity above parting surface is h = 300 mm

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So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

            $=2 \pi (0.23)^2$

            $=0.3324 \ m^2$

$F=\rho g A h$

   $=7200 \times 9.81 \times 0.3324 \times 0.3$

     = 7043.42 N

3 0
2 years ago
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