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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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10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
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Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

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3 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
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