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3 years ago

What is the wavelength of the wave

Physics

2 answers:

sladkih [1.3K]3 years ago

8 0

Explanation:

it is equal to the speed (v) of a wave train in a medium divided by its frequency (f): λ = v/f. Waves of different wavelengths.

Kazeer [188]3 years ago

5 0

Answer:

It is measured in the direction of the wave. Description: Wavelength is the distance from one crest to another, or from one trough to another, of a wave (which may be an electromagnetic wave, a sound wave, or any other wave).

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WHY is it so important to include units when expressing numbers?

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Answer:

Without units, the results are unclear and it is hard to keep track of what each seperate measurement entails.

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What type of objects cannot pull to magmets

boyakko [2]

Answer:

brass, copper, zinc and aluminum, wood, glass, and plastic

Explanation:

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3 years ago

Read 2 more answers

Comparing ultraviolet photons to radio photons, we can say that the ultraviolet photons have ______ energy, ______ frequency, an

guajiro [1.7K]

Answer:

Energy of ultra violet photons < Energy of radio waves

Wavelength of ultra violet photons > Wavelength of radio waves

Explanation:

The increasing order of the wavelength is given by

Gamma rays

x rays

Ultra violet rays

Visible radiations

Infrared rays

micro waves

Radio waves

So, ultra violet rays has larger wavelength than radio waves.

As we know that the energy is inversely proportional to the wavelength.

So, the energy of radio waves is more than the energy of ultra violet waves.

Energy of ultra violet photons < Energy of radio waves

Wavelength of ultra violet photons > Wavelength of radio waves

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3 years ago

You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference betwee

LUCKY_DIMON [66]

Answer:

In this scenario adding the dielectric material in between the plates will have no effect on the capacitance of the plates since the voltage remains unchanged

Explanation:

Normally Introducing a dielectric into a capacitor decreases the electric field, which decreases the voltage, which increases the capacitance.

A capacitor with a dielectric stores the same charge as one without a dielectric, but at a lower voltage.

Voltage and capacitance are inversely proportional when charge is constant.

Now in this case the voltage remains the same hence the charges remain the same also because voltage is inversely proportional to capacitance

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4 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

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4 0

3 years ago