Which two atmospheric gases do not react with many substances?

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

A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a

Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

$\sum F = ma$

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

$N-mg=ma$

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

$N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N$

So the closest answer is

D. 48.985 N

3 0

3 years ago

Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30ms

nexus9112 [7]

S= ?
U= 0 $ms^{-1}$
V= 30 $ms^{-1}$
A= 9.8 $ms^{-2}$
T=

$V^2 = U^2 + 2AS$
$900 = 19.6S$
$\frac{900}{19.6} = S$
S = 45.92 $m$

6 0

4 years ago

Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids i

777dan777 [17]

Answer:

its 45 over 6

Explanation:the answer is in the question

8 0

3 years ago

Read 2 more answers

Abox has a mass of 14.4 kg. The length is 4 m longthe width is 1 m and the height is 5 m. What is the density of the box?

valkas [14]

Answer:

0.72 kg per cubic m

Explanation:

Mass = 14.4 kg

Volume = lbh = 4*1*5 = 20 cubic m

$\because \: density \\ \\ = \frac{mass}{volume} \\ \\ = \frac{14.4}{20} \\ \\ = 0.72 \: kg {m}^{ - 3}$

3 0

3 years ago