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2 years ago

How large a band of frequencies does each television broadcasting channel get ?

Physics

2 answers:

k0ka [10]2 years ago

8 0

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

Anton [14]2 years ago

7 0

Answer:

Explanation:

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The wavelength of an electromagnetic waves is __________ also known as the period. the number of waves that pass a given point i

Andreas93 [3]

The sentence can be completed as follows:

The wavelength of an electromagnetic waves is the spatial distance between two successive troughs.

Note that the wavelength of a wave can also measured as the spatial distance between two successive crests of the wave. Also note that the second part of the sentence ("also known as the period") is not true, because period is another thing (in fact, the period is the time interval between two successive troughs).

6 0

2 years ago

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

<u>Gravity at a height h above the surface of the moon will be given as:</u>

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

2 years ago

A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = $\frac{energy}{time}$

The energy needed to light one bulb= $power*time$

Given time = 1minute = 60 seconds

Energy = $32W*60sec$ =1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = $1920*24$ =46080J=46.08KJ

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2 years ago

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Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a

Gnoma [55]

The distance covered by Arthur is
3t
The distance covered by Betty from where Arthur started is
100 - 2t
They meet each other when this distance is equal
3t = 100 - 2t
t = 20 s

Since the dog has a constant speed of 5m/s, the distance that Spot has run is
5 m/s (20s) = 100 m

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2 years ago

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Will mark brainlest helpp!!!!!!

Luda [366]

Answer:

I don't know if it is correct or not.

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2 years ago

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