Home Page Textbooks Physics: Principles and Problems Page 706: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 706: Practice Problems

Exercise 15

Step 1

1 of 2

Electromagnetic waves travel at constant speed, the speed of light, in air regardless of it is frequency

$$
c=3times 10^8frac{textrm{m}}{textrm{s}}
$$

Result

2 of 2

$$
c=3times 10^8frac{textrm{m}}{textrm{s}}
$$

Exercise 16

Step 1

1 of 2

In order to solve this problem we are going to use the frequency-wavelength relation

$$
c=lambda f
$$

from where we can express the the wavelength as

$$
lambda=frac{c}{f}=frac{3times 10^8}{5.7times 10^{14}}
$$

Finally, we have that

$$
boxed{lambda=5.26times 10^{-7}textrm{ m}}
$$

Result

2 of 2

$$
lambda=5.26times 10^{-7} textrm{ m}
$$

Exercise 17

Step 1

1 of 2

In order to solve this problem we are going to use the frequency-wavelength relation

$$
c=lambda f
$$

from where we can express the the wavelength as

$$
lambda=frac{c}{f}=frac{3times 10^8}{8.2times 10^{14}}
$$

Finally, we have that

$$
boxed{lambda=3.7times 10^{-7}textrm{ m}}
$$

Result

2 of 2

$$
lambda=3.7times 10^{-7}textrm{ m}
$$

Exercise 18

Step 1

1 of 2

In order to solve this problem we are going to use the frequency-wavelength relation

$$
c=lambda f
$$

from where we can express the frequency as

$$
f=frac{c}{lambda}=frac{3times 10^8}{2.2times 10^{-2}}
$$

Finally, we have that

$$
boxed{f=1.4times 10^{10}textrm{ Hz}}
$$

Result

2 of 2

$$
f=1.4times 10^{10}textrm{ Hz}
$$

Exercise 19

Step 1

1 of 2

In order to solve this problem, we are going to use the definition of the speed of an electromagnetic wave in an isotropic medium (dielectric tensor=const)

$$
v=frac{c}{sqrt{K}}=frac{299,792,458}{sqrt{1.00054}}
$$

Which gives that in air the speed of electromagnetic waves is

$$
boxed{v=299711547, frac{textrm{m}}{textrm{s}}}
$$

Result

2 of 2

$$
v=299711547, frac{textrm{m}}{textrm{s}}
$$

Exercise 20

Step 1

1 of 2

In order to solve this problem, we are going to use the definition of the speed of an electromagnetic wave in an isotropic medium (dielectric tensor=const)

$$
v=frac{c}{sqrt{K}}=frac{3times 10^8}{sqrt{1.77}}
$$

Which gives that in water the speed of electromagnetic waves is

$$
boxed{v=2.25times 10^8, frac{textrm{m}}{textrm{s}}}
$$

Result

2 of 2

$$
v=2.25times 10^8, frac{textrm{m}}{textrm{s}}
$$

Exercise 21

Step 1

1 of 2

In order to solve this problem, we are going to use the definition of the speed of an electromagnetic wave in an isotropic medium (dielectric tensor=const)

$$
v=frac{c}{sqrt{K}}
$$

From here one can express the dielectric constant as

$$
K=frac{c^2}{v^2}
$$

which after we plug in the values becomes

$$
K=frac{3^2times 10^{16}}{2.43^2 times 10^{16}}
$$

$$
boxed{K=1.52}
$$

Result

2 of 2

$$
K=1.52
$$

Exercise 22

Step 1

1 of 2

The propagation of electromagnetic waves is based on interchangeable nature of the electric and magnetic field which regenerate each other in the process thus causing the propagation.

Result

2 of 2

The electric and magnetic field change with time re-inducing each other along the way.

Exercise 23

Step 1

1 of 2

Electromagnetic waves have all the properties of the well-known mechanical waves such as wavelength, frequency, amplitude, etc. However, they have one fundamental property that makes them unique in nature and that is that they do not need a medium to travel. They can travel in a vacuum, too. Another distinctive feature is polarization, i.e. the can be polarized along different spatial directions.

Result

2 of 2

Electromagnetic waves do not require a medium for propagation and they can be polarized.

Exercise 24

Step 1

1 of 2

In order to solve this problem we are going to use the frequency-wavelength relation

$$
c=lambda f
$$

from where we can express the frequency as

$$
f=frac{c}{lambda}=frac{3times 10^8}{1.5times 10^{-2}}
$$

Finally, we have that

$$
boxed{f=2times 10^{13}textrm{ Hz}}
$$

Result

2 of 2

$$
f=2times 10^{13}textrm{ Hz}
$$

Exercise 25

Step 1

1 of 2

Since the electric field should go along the conductor i.e. rods, it has to be horizontal, too.

Result

2 of 2

It is horizontal.

Exercise 26

Step 1

1 of 2

This is a necessity since the dish’s antenna acts as a collector or as a lens which collected waves focuses to its focal point. This condition limits dish antennas to a narrow range of angles capable to receive the incoming signals well and with a sufficient strength.

Result

2 of 2

If not aligned well with the transmitter the signal strength is not going to be maximal.

Exercise 27

Step 1

1 of 2

In order to answer this question we have to understand that the frequency is inversely proportional to the wavelength i.e.

$$
fpropto frac{1}{lambda}
$$

This means that the higher frequency waves will have shorter wavelengths and vice versa. Therefore the channels from 2 to 6 will require longer antennas.

Result

2 of 2

The channels 2-6 will demand longer antennas.

Exercise 28

Step 1

1 of 2

In order to solve this problem, we are going to use the definition of the speed of an electromagnetic wave in an isotropic medium (dielectric tensor=const)

$$
v=frac{c}{sqrt{K}}
$$

From here one can express the dielectric constant as

$$
K=frac{c^2}{v^2}
$$

which after we plug in the values becomes

$$
K=frac{3^2times 10^{16}}{1.98^2 times 10^{16}}
$$

$$
boxed{K=2.3}
$$

Result

2 of 2

$$
K=2.3
$$

Exercise 29

Step 1

1 of 2

The ozone layer is protecting the life on Earth from the harmful UV radiation. Although not ionizing, UV light can damage chemical structure of the skin in human and it can kill many microorganism which could damage the biosphere as total.

Result

2 of 2

The UV is not ionizing radiation but it can cause the chemical changes on the skin.

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Page 706: Practice Problems

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