If the length of one side of a square is 12.0 m, what is the perimeter of the square?

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

3 years ago

Please help help me please

vodomira [7]

The answer is either c or d but c is the best answer

4 0

3 years ago

Which of the following is a small electronic component made up of transistors (tiny switches) and other miniaturized parts?

Vlad [161]

Answer:

Integrated circuit (IC)

Explanation:

An integrated circuit ( IC ) is a semiconductor which contains multiple electronic components

interconnected to form a complete electronic function. Integrated circuits are the most essential part of all electronic products.

Modern integrated circuits contain as much as billions of circuit components such as transistors , diodes , resistors , and capacitors

onto a single monolithic die .

5 0

3 years ago

What is sludge dumping?<br>HELP ASAP PLEASE

yulyashka [42]

Answer:

What is sludge dumping? Sludge is the solid waste in raw sewage. Sludge dumping is discharging that waste into the ocean.

Explanation:

5 0

3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$

where we have

E = 310 V/m is the amplitude of the electric field

$\mu_0$ is the vacuum permeability

c is the speed of light

Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

(g) $1.53\cdot 10^{-8} kg m/s$

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

(h) $8.47\cdot 10^{-7} N/m^2$

For a surface that totally absorbs the wave, the radiation pressure is given by

$p=\frac{}{c}$

where we have

$=254.2 W/m^2$

$c=3\cdot 10^8 m/s$

Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

8 0

3 years ago