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3 years ago

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

2 answers:

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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A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total

Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

$v_2 = v - 1$

now we know that total time of the motion is 19 hr

so we will have

$t_1 = \frac{90}{v}$

$t_2 = \frac{90}{v - 1}$

so we will have

$t_1 + t_2 = 19 hr$

$\frac{90}{v} + \frac{90}{v-1} = 19$

$90(2v - 1) = 19(v^2 - v)$

$19 v^2 - 199 v + 90 = 0$

by solving above equation we have

$v = 10 mph$

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0

3 years ago

An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp

Natali5045456 [20]

Answer:

h = 20 m

Explanation:

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or $9.8 m/s^2$ .

The weight of an object of mass m is:

W = m.g

Thus, the GPE is:

U=W.h

Solving for h:

$\displaystyle h=\frac{U}{W}$

The weight of the owl is W=40 N and its GPE is U=800 J.

$\displaystyle h=\frac{800}{40}=20$

h = 20 m

3 0

3 years ago

A home run is hit in such a way that the baseball just clears a wall 21m hgh located 130m from home plate the ball is hit at an

svlad2 [7]

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

Download pdf

3 0

3 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

6 0

3 years ago

Read 2 more answers

The wavelength of light is 5000 angstrom. Express it in nm and m.

Ierofanga [76]

Answer:

1 angstrom = 0.1nm

5000 angstrom = 5000/1 × 0.1nm

$1 \: angstrom = 1 \times {10}^{ - 10} m$

5000 angstrom = 5000 × 1 × 10^-10

Hope this helps you

7 0

3 years ago