All categories

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

You might be interested in

What kind of symmetry do you have

Keith_Richards [23]

During the daytime, I have mostly line symmetry.

During the night, I often have almost spherical symmetry.

5 0

3 years ago

Read 2 more answers

Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)

8 0

3 years ago

Read 2 more answers

Gravity __________ your kinetic energy when you are driving uphill and __________it when you are driving downhill.

Oxana [17]

Gravity decreases your kinetic energy when you are driving uphill since the direction of motion is opposite for both. Driving uphill is force going upward while gravity pulls object down. When it is going downhill, the car tends to go faster since the gravity helps the object to go down by adding another value to the total acceleration of the motion of the object. Using the forces of balance, an object going up tends to become heavier while object going down tends to become lighter because of the gravity factor. Another analogy is the motion of elevators going up and down that incurs effects to your weiight.

3 0

3 years ago

A 90-kilogram physics student would weigh 2970 Newtons on the surface of planet X. What is the magnitude of the acceleration due

KiRa [710]

<u>Weight = (mass) x (acceleration of gravity)</u>

Divide each side by (mass),and we have

Acceleration of gravity = (weight) / (mass)

Acceleration of gravity = 2,970/90 = 33 newtons per kilogram = <em>33 m/s²</em>

5 0

3 years ago

4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2

bazaltina [42]

Answer:

$1030.83\ \text{N}$

Explanation:

$\Delta L$ = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = $\dfrac{\pi}{4}d^2$

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

$\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}$

La tensión en la cuerda es $1030.84\ \text{N}$ .

8 0

3 years ago