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Anon25 [30]
3 years ago
14

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:__________.
Physics
1 answer:
mestny [16]3 years ago
8 0

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.

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FrozenT [24]

Answer:

7.5 km

Explanation:

h5 per hour means that he traveled 5 km in 1 our. And then half of the hour, which means half an hour 5 km which is 2.5.

5 + 2.5 = 7.5

or just 1.5 x 5 = 7.5

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If a ball is dropped from rest and falls 8 m to the ground, what is the speed just before it
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Answer:

12.5m/s

Explanation:

(Assuming the question was asking for the speed just before it hit the ground)

We can use the first key equation of accelerated motion

Vf^2 = Vi^2+2aΔd

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How did putting myths in writing help develop philosophy?
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Answer: A

Explanation:

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3 years ago
g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Fin
schepotkina [342]

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

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3 years ago
An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to
Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

\theta_1 is the angle of incidence (in the first medium)

\theta_2 is the angle of refraction (in the second medium)

Let's now consider a situation in which

n_1 > n_2

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

Where \frac{n_1}{n_2} is a number greater than 1. This means that above a certain value of the angle of incidence \theta_1, the term on the right can become greater than 1. So this would mean

sin \theta_2 > 1

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

\theta_c =sin^{-1}(\frac{n_2}{n_1})

8 0
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