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melamori03 [73]
4 years ago
10

If the wave being sent is transmitted via an electromagnetic wave explain how the digital signal would be perceived if the signa

l was sent using:
- Amplitude modulation
&
- Frequency modulation

Will mark brainliest, plz help.
Physics
1 answer:
Andrej [43]4 years ago
6 0

Answer:

amplitude is commonly for transmitting messages with a radio carrier wave, the amplitude (signal strength) of the carrier wave is varied in proportion to that of the message signal. at the receiving end, the message signal is extracted from the modulated carrier by demodulation. frequency is the encoding of information in a carrier wave with instantaneous frequency. with digital data, the frequency of the carrier is shifted among a set of frequencies, using digits like 1 and 0

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Does a light bulb at a temperature of 2500K produce as white a light as the sun at 6000K
sattari [20]

Answer:

No. ... UV light causes sunburn, whereas visible light does not.

Explanation:

7 0
3 years ago
1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.
AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}

Where;

v = The image distance from the lens

We get;

v = \dfrac{u \times f}{u - f}

Therefore;

v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0
3 years ago
In still​ water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles ​downstream,
Vladimir79 [104]
<span>The current is 6 miles per hour.
   Let's create a few equations:
 Traveling with the current:
 (18 + c)*t = 16

   Traveling against the current:
 (18 - c)*t = 8

   Let's multiply the 2nd equation by 2
 (18 - c)*t*2 = 16

   Now subtract the 1st equation from the equation we just doubled.
 (18 - c)*t*2 = 16
 (18 + c)*t = 16

   (18 - c)*t*2 - (18 + c)*t = 0
 Divide both sides by t
 (18 - c)*2 - (18 + c) = 0

   Now solve for c
 (18 - c)*2 - (18 + c) = 0
 36 - 2c - 18 - c = 0
 36 - 2c - 18 - c = 0
 18 - 3c = 0
 18 = 3c
  6 = c

   So the current is 6 mph.
   Let's verify that.
 (18 + 6)*t = 16
 24*t = 16
 t = 16/24 = 2/3

   (18 - 6)*t = 8
 12*t = 8
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   And it's verified.</span>
4 0
3 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
lbvjy [14]
-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s 

                                                                           = <span> 3,600 meters .</span>
___________________________________

The formula that has all of this in it is the formula for 
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is 
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds. 

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !  

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with. 
5 0
3 years ago
How many Joules in 41.87 kcal?
gladu [14]
The answer is 175184.08 joules
4 0
3 years ago
Read 2 more answers
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