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3 years ago

What affect, if any, does increasing the speed of the plunger have on the wavelength of the waves

1 answer:

6 0

As the speed of the plunger increases, the wavelength of the waves decreases. The greater the frequency, the smaller the wavelength. The smaller the frequency, the greater the wavelength. When we increase the speed of the plunger, the frequency of the waves also increases, and just like with the size of the ball, it’s the speed of the plunger and the frequency of the waves are directly related.

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A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

The force of friction always opposes the _______.

Ierofanga [76]

Answer:

motion

Explanation:

uh look it up?

8 0

3 years ago

Read 2 more answers

The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer

zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = $5.0\ m/s^2$

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

$F = ma$

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in $m/s^2$

Now, we substitute m = 20 kg and a = $5.0\ m/s^2$ in Newton's second law,

$F = 20\times 5.0$

$F = 100\ m/s^2$

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 0

3 years ago

A natural force of attraction exerted by the earth upon objects, that pulls

Mashutka [201]

A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .

6 0

3 years ago

A force of 10 n acts on a mass of 5 kg producing an acceleration of 1.5 m/s2 what is the magnitude of the force of friction acti

prisoha [69]

The friction is 2.5N. The Net force is 10 N - 2.5 N .= 7.5 N.
acceleration = 7.5 / 5 = 1.5 m/s^2

8 0

3 years ago