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sergij07 [2.7K]
3 years ago
10

Which of the following frequencies falls in the range of RF waves used by commercial radio broadcasting stations?

Physics
2 answers:
erma4kov [3.2K]3 years ago
7 0

A).  600,000 Hz  or  600 KHz
Yes.  Commercial broadcasters operate here.
This is the '600' on your AM radio dial.

B).  60 Hz
No.  In principle, this frequency might be used for communication or
commercial broadcasting, but it suffers from two inconvenient truths:
-- An efficient antenna for 60 Hz ... either transmitting or receiving ...
needs to be almost 780 miles long.
-- This is the frequency of the electric power utility in the US and
Canada, so every outlet, wire, cable, lamp cord, and electric line
on a pole RADIATES a little bit of signal at this frequency.  That's
an awful lot of interference.

C).  6,000,000 Hz  or  6 MHz
There's a lot of broadcasting activity here, but it's not commercial
music, news, and sports into local homes and cars. 
It's foreign short-wave broadcast, bringing news, propaganda, and
culture from one country to another.  Pretty interesting to browse.

D).  6,000 Hz  or  6 KHz.
No.  Not used for communication, for an interesting reason:
This frequency is smack in the middle of the human hearing range.
So if it were used for communication ... with high-power transmitters
here and there ... then you wouldn't hear it in the air.  But wherever
wires were being used to carry sound ... your stereo's speaker wires,
wires from your player to your ear-buds, wires to the telephones in
your house etc ... the wires would act as antennas, picking up 
broadcasts at 6 KHz, and the broadcasts would get into everything.
Not a smart plan.

creativ13 [48]3 years ago
4 0

On page 103 it explains that The radio frequencies used by  commercial radio broadcasting stations range from about 550,000 Hz to 1,700,000 Hz. So the correct answer is A. 600,000 Hz

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Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On
8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

4 0
3 years ago
You use a lever to lift a heavy tree branch you apply a force of 50 n and the lever lifts the branch
valentinak56 [21]

1.8 is the mechanical advantage of the lever.

<h3>Definition of mechanical advantage</h3>

The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.

The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.

Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.

In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.

Mechanical advantage= 90/50 =1.8

Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?

To learn more about the Mechanical advantage visit the link

brainly.com/question/16617083

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5 0
1 year ago
The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0
3 years ago
Read 2 more answers
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0
3 years ago
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
dedylja [7]

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

Speed of sound wave,v=1522 m/s

We have to find the period of the wave.

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

Using identity:\frac{a^x}{a^y}=a^{x-y}

Hence, the time period of the wave=2.2\times 10^{-5} s

4 0
3 years ago
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