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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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What type of electromagnetic wave does a light bulb and a radio antenna release?
andrew-mc [135]
I believe that a light bulb releases visible light and a radio antenna releases a radio waves
4 0
3 years ago
On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
lys-0071 [83]

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring [email protected] = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

                                 T  = 2*pi*sqrt(k / m )    ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

                                 F_net = k*x - m*g_x = 0

                                 (k / m) = (g_x / x)    .... 2

- substitute Eq 2 into Eq 1:

                                  2*pi / T = sqrt ( g_x / x )

                                   g_x = (2*pi / T )^2 * x

- Evaluate g_x:

                                  g_x = (2*pi / (19 / 11) )^2 * 0.226

                                  g_x = 3.0 m / s^2

                                 

                       

3 0
3 years ago
An object is moving at a constant speed along a straight line. Which of the following statements is not true? A. There must be a
Svetradugi [14.3K]

Answer:

False statement = There must be a non-zero net force acting on the object.  

Explanation:

An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.

a=\dfrac{dv}{dt}

a = 0

⇒ The acceleration of the object is zero.

The product of force and acceleration gives the magnitude of force acting on the object i.e.

F = m a = 0

⇒  The net force acting on the object must be zero.

So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.

6 0
3 years ago
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8 0
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Read 2 more answers
During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.

4 0
3 years ago
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