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Mazyrski [523]
3 years ago
10

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 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.
Physics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

b.) All these technologies use radio waves, including 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.

Explanation:

For option D. The frequency range of microwave ovens is 2450 MHz = 2.4 GHz, which intersects with wireless internet technology with range of 2.4 to 2.6 GHz.

For E, wavelenght and frequency are inversely proportional. Wireless internet service has the greatest frequency band and hence the shortest wavelenght band.

For F, in all these radiations, the highest Freq is 2.6 GHz and the lowest is 40 MHz. Wavelenght is speed of light (3x10^8 m/s) divided by the frequency.

2.6 GHz = 2.6x10^9 Hz

Wavelenght = 3x10^8 ÷ 2.6x10^9 = 0.1 m

40 MHz = 40x10^6

Wavelenght = 3x10^8 ÷ 40x10^6 = 7.5 m

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Answer:

the effects that a jet and the magnetic fields have on a ProStar Is :

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To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1
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Answer:

2.2\times 10^{-5} s

Explanation:

We are given that  

The wavelength of sound wave=\lambda=3.3 cm=3.3\times 10^{-2}m/s

1 cm/s=10^{-2}m/s

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We know that

Frequency=\nu=\frac{v}{\lambda}

Using the formula

Frequency =\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4} Hz

Time period=\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s

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3 years ago
Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
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This question can be solved from the Kepler's law of planetary motion.

As per this law the square of time period of a planet  is proportional to the cube of semi major axis.

Mathematically it can be written as   T^{2} \alpha R^{3}

                                                          ⇒T^{2} = KR^{3}

Here K is the proportionality constant.

If T_{1} andT_{2} are the orbital periods of the planets and

R_{1} and R_{2} are the distance of the planets from the sun, then Kepler's law can be written as-

          \frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }

      ⇒ R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

  Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.

Let the distance from sun and orbital period of Saturn is denoted as R_{1} and T_{1} respectively.

Let the distance  from sun and orbital period of earth is denoted as R_{2} and T_{2} respectively.

we are given thatT_{1} =29.46 years

we know that R_{2} = 1 AU and T_{2} = 1 year.

1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.

Hence distance of Saturn from sun  is calculated as -

From Kepler's law as mentioned above-

                                    R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

                                             =[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU

                                    =867.8916 AU^{3}

                                        ⇒R_{1} =\sqrt[3]{867.8916}

                                           =9.5386 AU [ans]

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