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Furkat [3]
3 years ago
15

The ratio of the wavelength of AM radio waves traveling in a vacuum to the wavelength of FM radio waves traveling in a vacuum is

approximately
1.) 1 to 1
2.) 2 to 1
3.) 10^2 to 1
4.) 10^8 to 1
Physics
1 answer:
Debora [2.8K]3 years ago
3 0

This question is a big fat non sequitur !

The wavelength of radio waves traveling through vacuum only depends on the frequency that the radio station is licensed to broadcast on, (which had better be the frequency of the transmitter that they buy and use, or they're in big trouble).

The wavelength does NOT depend on the type of modulation that's used to put information onto the signal.  

An amateur radio (ham) operator may very well start out using FM to talk over his radio to somebody else, and then for some reason they may decide to switch to AM.  They can do that without ANY change in the wavelength of their transmissions.

Now, in the USA and many other countries, it so happens that all AM stations are licensed by their governments to transmit their programs on a channel somewhere between 500 KHz and 1.6 MHz, and all FM stations are licensed by their governments to transmit their programs on a channel somewhere between 88 MHz and 108 MHz.  (And THAT's what the radio receivers in these countries are built to receive.)

Then we might say that all of the AM stations are grouped around 1 MHz, and all of the FM stations are grouped around 100 MHz.  The FM frequencies are very roughly 100 times the AM frequencies, so the AM wavelengths are very roughly  100 times the FM wavelengths.  That's <em>choice (3)</em> .

But please don't get the idea that it has anything to do with using AM or FM technology.  It's just a matter of where in the spectrum the government decided to put the AM stations and where they put the FM stations.

For that matter . . . An analog TV station uses an AM signal for the picture and an FM signal for the sound, and it all goes in the same channel, with just about the same wavelengths !

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Answer:

4) Increase

Explanation:

Hope this helps!

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4 0
4 years ago
Which statement best describes beaches?
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6 0
3 years ago
Read 2 more answers
Suppose humans have weights which are normally distributed with mean 170 lbs and SD 50 lbs. If 400 humans are selected at random
MrRa [10]

Answer:

 P(x< 175)= 0.9772

Explanation:

given,

mean weight of human = μ = 170 lbs

standard deviation = SD = 50 lbs

N = 400 humans

By using central limit theorem,

P (x< 175)

P(x< 175)= P(z

P(x< 175)= P(z

P (x< 175)= P(z

P (x< 175)= P(z

using z-table

 P (x< 175)= 0.9772

hence, the probability that total weight is less tan 175 lbs is equal to

 P(x< 175)= 0.9772

7 0
3 years ago
A single slit is illuminated by light of wavelengths λa and λb, chosen so that the first diffraction minimum of the λa component
agasfer [191]

Answer:

λ_A = 700 nm ,   m_B = m_a 2

Explanation:

The expression that describes the diffraction phenomenon is

         a sin θ = m λ

where a is the width of the slit, lam the wavelength and m an integer that writes the order of diffraction

a) They tell us that now lal_ A m = 1

         a sin θ = λ_A

coincidentally_be m = 2

          a sin  θ = m λ_b

as the two match we can match

         λ _A = 2 λ _B

         λ_A = 2 350 nm

         λ_A = 700 nm

b)

For lam_B

       a sin  λ_A  = m_B  λ_B

For lam_A

        a sin θ_A = m_ λ_ A

to match they must have the same angle, so we can equal

           m_B  λ_B = m_A  λ_A

           m_B = m_A  λ_A / λ_B

           m_b = m_a 700/350

           m_B = m_a 2

8 0
3 years ago
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