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3 years ago

The flow of electrons through wires and components is known as:

Physics

1 answer:

Dvinal [7]3 years ago

3 0

Answer:

Electric Current

Explanation:

An electric current is defined as the flow of electrons through the wire and the components in a circuit.

This flow of electrons is caused by the presence of a battery, which creates a potential difference in the circuit. As a result of this potential difference, the electrons in the wire flow from the negative pole of the battery towards the positive pole (passing through the wire and the components of the circuit).

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A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge

Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is $\lambda$ and the net charge per unit length is $2 \lambda$ .

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let $\lambda_i\ and\ \lambda_o$ are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

$\phi=\dfrac{q_{enclosed}}{\epsilon_o}$

$E.A=\dfrac{q_{enclosed}}{\epsilon_o}$

$0=\dfrac{\lambda_i+\lambda}{\epsilon_o}$

$\lambda_i=-\lambda$

For outer surface,

$\lambda_i+\lambda_o=2\lambda$

$-\lambda+\lambda_o=2\lambda$

$\lambda_o=3\lambda$

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

$E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}$

$E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}$

Hence, this is the required solution.

6 0

3 years ago

Which image illustrates refraction?

maks197457 [2]

Answer:

B illustrates refraction

3 0

3 years ago

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.

AleksAgata [21]

Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

Explanation:

The frequency of the $n^{th}$ harmonic of a vibrating string of length L, linear density $\mu$ under a tension T is given by the formula:

$f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}$

a) So for the fundamental mode (n=1) we have, substituting our values:

$f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz$

b) The frequency difference between successive modes is the fundamental frequency, since:

$f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz$

3 0

2 years ago

What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

4 0

3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago