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DENIUS [597]
2 years ago
11

How are electromagnetic waves used in a bar-code scanner?

Physics
1 answer:
meriva2 years ago
5 0

The electromagnetic waves used in a bar-code scanner by  an image is produced when a silicon chip releases electrons by the photoelectric effect and  Visible light is reflected off an object and then captured by a charge-coupled device.

<h3>What is electromagnetic wave?</h3>

The electromagnetic wave is the white light or Sunlight coming directly from the Sun. This light ray is the form of wave. This light when falls on any object segregates into the range of all types of EM radiation.

Radiation is packet of energy that travels and spreads out. It has radio waves, infrared waves and visible light waves, ultraviolet rays and microwaves.

Bar code scanner uses the photoelectric effect and visible light reflection phenomenon.

Thus, the correct options are A and C.

Learn more about electromagnetic wave.

brainly.com/question/3101711

#SPJ1

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How long will it take an object to hit the ground if dropped from 100 meters
mash [69]

Answer:

The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.

Good luck!

Explanation:

8 0
3 years ago
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A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

8 0
2 years ago
The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u  = 25 cm

image distance   v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f  = .02215

f = 45.15 cm .

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Which of the following had have the highest viscosity ?
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