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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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1 year ago
In the following figure, if AB ǁ CD, then find the measure of PCD and CPD.​
mina [271]

Answer:

CPD = 80

PCD = 44

Explanation:

Given

AB || CD

BAD = 56

CPA = 100

See attachment

Required

Determine PCD and CPD

First, we need to calculate CPD

Since DPA is a straight line and CPA = 100;

We have that:

CPA + CPD = 180 --- angle on a straight theorem

Substitute 100 for CPA

100 + CPD = 180

Subtract 100 from both sides

100-100 + CPD = 180-100

CPD = 80

Next, we calculate PCD

We have that:

DAB= ADC = 56  --alternate angle

In triangle PCD

PCD + CPD + PDC = 180 --- angles in a triangle

Where

PDC = ADC = 56

So, we have:

PCD +80 + 56 = 180

PCD +136 = 180

Subtract 136 from both sides

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6 0
3 years ago
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6 0
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A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
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Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
2 years ago
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