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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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What will be the ME of a machine that produces a 240 N work with a 300

Elden [556K]

Answer:

Efficiency = 80%

Explanation:

Given the following data;

Work output = 240 N

Work Input = 300 N

To find the mechanical efficiency of a machine;

$Efficiency = \frac {Out-put \; work}{In-put \; work} * 100$

Substituting into the equation, we have;

$Efficiency = \frac {240}{300} * 100$

$Efficiency = 0.8 * 100$

Efficiency = 80%

Therefore, the mechanical efficiency of the machine is 80 percent.

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3 years ago

Can anyone plz answer this now and give me a right answer?

klemol [59]

Answer:

90 degree hope it help :))

5 0

2 years ago

Read 2 more answers

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)