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dangina [55]
3 years ago
13

The critical angle for a substance is measured at 53.7 degrees. Light enters from air at 45.0 degrees. At what angle it will con

tinue to travel?
Physics
1 answer:
lys-0071 [83]3 years ago
8 0

Answer:

34.74

Explanation:

Critical angle is represented by the formula

θ(c) = sin^-1 (n2/n1)

And we have been given that the critical angle is 53.7, so we substitute

53.7 = sin^-1 (n2/n1), rearranging, we have

Sin 53.7 = n2/n1

0.8059 = n2/n1

Again, we know that the second medium is air, and the refractive index of air is 1, so

0.8059 = 1/n1, making n1 subject of formula

n1 = 1/0.8059

n1 = 1.24

Now, we have gotten both indices, using Snell's law of refraction, we then find the needed refraction

Sin θ(r) = n(i)/n(r) Sin θ(i)

Now, solving, we have

Sin θ(r) = 1/1.24 * sin 45

Sin θ(r) = 0.8059 * 0.7071

Sin θ(r) = 0.5699

θ(r) = Sin^-1 0.5699

θ(r) = 34.74

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The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies
mestny [16]

Complete question is;

The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:

garage door openers: 40.0 MHz

standard cordless phones: 40.0 to 50.0 MHz

baby monitors: 49.0 MHz

FM radio stations: 88.0 to 108 MHz

cell phones: 800 to 900 MHz

Global Positioning System: 1227 to 1575 MHz

microwave ovens: 2450 MHz

wireless internet technology: 2.4 to 2.6 GHz

Which of the following statements correctly describe the various applications listed above? Check all that apply.

a.) All these technologies use radio waves, including low-frequency microwaves.

b.) All these technologies use radio waves, including high-frequency microwaves.

c.) All these technologies use a combination of infrared waves and high-frequency microwaves.

d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.

e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.

f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.

g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.

Answer:

B, D, E, F are the correct statements.

Explanation:

Looking at the options;

A) This is true because radio waves are electromagnetic radiation being used today in television, mobile phones, radios and other areas of communication technologies. And the examples given to us fall in the category of technologies that use radio waves.

B) microwaves usually have long wavelengths and low frequencies. However, sometimes they could have high frequencies usually more than radio waves. Thus, this option is correct.

C) This option is wrong because it's not all the listed technologies that use combination of infrared waves and high-frequency microwaves.

D) we are given the frequency of microwave ovens as 2450 MHz.

Converting to GHz gives; 2.45 GHz.

We are told that wireless internet technology has frequency between 2.4 to 2.6 GHz. Thus, microwave frequency falls in the same range as wireless internet technology and thus the statement is true.

E) we know that wavelength is inversely proportional to frequency. This means that the higher the frequency, the shorter the wavelength.

In the frequencies given to us, wireless internet technology have the highest frequency which means they have the shortest wavelength. The statement is true.

F) from the frequencies given to us, the smallest is garage door openers = 40.0 MHz = 40 × 10^(6) Hz while the biggest is 2.6 GHz = 2.6 × 10^(9) Hz

Formula for wavelength is;

Wavelength = speed of light/frequency

Speed of light = 3 × 10^(8) m/s

Thus;

Wavelength = (3 × 10^(8))/(40 × 10^(6))

Or wavelength = (3 × 10^(8))/(2.6 × 10^(9))

So,wavelength = 7.5 m or 0.12 m

This falls into the given range of 0.10 to 10.0 m.

Thus, the statement is true.

8 0
3 years ago
What is the difference between the formation of an ionic bond and formation of a covalent bond?
Vikentia [17]

Answer:

Answer to the question:

Explanation:

Differences between ionic bond and covalent bond:

The ionic bond occurs between two different atoms (metallic and non-metallic), while the covalent bond occurs between two equal atoms (non-metallic).

In the covalent bond there is an electron compartment, while in the ionic bond there is an electron transfer.

Ionic bonds have a high melting and boiling point, while covalent bonds usually have a low point.

5 0
3 years ago
What is the name for family labeled #4 (Yellow)?
goldfiish [28.3K]

Answer:

transition metals im sorry if this was too late

4 0
3 years ago
A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?
Ivanshal [37]
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration g=9.81 m/s^2. Calling h its height at t=0, the height at time t is given by
h(t)=h- \frac{1}{2}gt^2
We are told thatn when t=0.75 s the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
0=h- \frac{1}{2}gt^2
h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m

2) The speed of the grapefruit at time t is given by
v(t)=v_0 +gt
where v_0=0 is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s
3 0
4 years ago
A ball is dropped from a height of 20m and renounce with the velocity which is 3/4 of d velocity with which it hits the ground.
Mashutka [201]

Answer:

1.73 seconds

Explanation:

The velocity the ball first hits the ground with is:

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)

v = -20 m/s

The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.

The time it takes to return to the ground is:

Δx = v₀ t + ½ at²

0 = (15 m/s) t + ½ (-10 m/s²) t²

0 = t (15 − 5t²)

t = √3

t ≈ 1.73 seconds

3 0
3 years ago
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