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3 years ago

Explain why radio and television stations use different frequencies to broadcast programs.

Physics

2 answers:

Sedaia [141]3 years ago

5 0

I'm not sure exactly what you're asking.

It could be

==> Why do radio stations use different frequencies from TV stations ?

or it could be

==> Why don't all radio stations, or all TV stations, use the same frequency ?

Radio and TV can't coexist among each other in the same "band"

of frequencies, because they use different amounts of "space" on

the dial. One analog TV channel uses enough dial space for about

600 AM radio stations, or 30 FM radio stations ! That's one click on

the TV channel knob !

So if they were all jumbled up together on the same dial and you

wanted to tune your radio from one AM station to another, you

might have to crank through enough space for 600 radio stations ...

or even 1200 or 1800 of them ... to go to the AM signal you want.

And maybe even worse than that ! I'm sure you've never heard what

a TV signal SOUNDS like on a radio. It is horrendous, and it is loud !

It sounds like a thousand cats shrieking at each other, and it never stops.

That's another good reason to move the TV transmissions to frequencies

where radios will never hear them. If radios just randomly tuned in to a

TV picture signal every now and then, a lot of people would be shocked

out of their socks. They would stop listening to radio, and thousands of

advertisers would not like that.

For the second question ...

OK, so we don't mix radio and TV in the same band of frequencies.

But why does each station need its own frequency ? Why not just

put every radio station on one frequency, and every TV station on

a single frequency that's different from the radio frequency ?

The answer is: It's because people don't want to listen to two radio

stations at the same time, or watch two TV movies at the same time.

We like to make our choice, and then watch them or listen to them

one at a time. And FREQUENCY is the only way our radios and TVs

know how to pick out ONE and ignore all the others.

If there are two, or 5 or 10 stations all on the same frequency within

10 or 20 miles from you, then when you tune your radio to that frequency,

you HEAR two, or 5 or 10, songs, church services, newscasts, political

speeches, or commercials, all at the same time.

So if all radio stations were on the same frequency, or all TV stations

were all on the same frequency, then any time you turned on your

radio or TV, you'd see or hear all of them together. Radio and TV

would completely lose their entertainment value, everybody would

give up watching and listening, and once again ... thousands of

advertisers would not like that.

After all, advertising is the main reason why we have so much radio and TV at all. The advertiser buys, the broadcaster sells, and YOUR eyes and ears are the product.

Alex73 [517]3 years ago

3 0

Interference is avoided when different frequencies are used by radio and television stations.

Interference may degrade the quality of the signals.

Radio and television stations use different frequencies to broadcast programs to avoid interference. Interference may degrade the quality of signals.

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3 years ago

Read 2 more answers

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$