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2 years ago

What are conductors, semiconductors, non-conductors, dielectrics?

Physics

1 answer:

elena55 [62]2 years ago

5 0

Conductors are substances that pass an electrical charge.

Semiconductors are substances whose electrical conductivity is lower than that of metals and greater than that of dielectrics.

Electricity nonconductors or insulators - in the terminology of Faraday - dielectrics (see). N. perfect does not exist; they represent only a large resistance to galvanic current and then different bodies in varying degrees (see Galvanic current), so that between poor and good conductors there are many bodies of average conductivity. N. The galvanic current is also the best insulators of static electricity. N. Heat or its bad conductors are at the same time electrical insulators (see Thermal Conductivity).

A dielectric (insulator) is a substance that is poorly conducting or not conducting at all. The concentration of free charge carriers in a dielectric does not exceed 108 cm-3. The main property of the dielectric is the ability to polarize in an external electric field. From the point of view of the band theory of a solid body, a dielectric is a substance with a band gap greater than 3 eV.

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Answer:

8.162

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I've had several mental breakdowns please help me

sesenic [268]

Answer:

The efficiency of the system is 63.7 %

Explanation:

Given;

input power of the motor, = 1.5 kW = 1,500 W

mass of the car lifted, m = 1300 kg

height through which the car was lifted, h = 1.8 m

time, t = 24 s

The output power of the motor is calculated as;

Output Power = F x v

= (mg) x (d/t)

= (1300 x 9.8) x (1.8 / 24)

= 12,740 x 0.075

= 955.5 W

The efficiency of the system is calculated as;

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The correct answer is 63.7%

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2 years ago

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

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2 years ago

Example of nuclear fission and nuclear fusion

EleoNora [17]

Answer:

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Explanation:

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2 years ago