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Dominik [7]
3 years ago
14

A barcode can be printed on a product to help identify the product. A barcode usually consists of several alternating vertical b

lack and white lines of various widths. A barcode scanner is used to read the information stored in a barcode. The barcode scanner projects a beam of light on the barcode. The light that reflects from the barcode is decoded to give information about the product.
Which of the following processes enables the scanner to decode the information from the barcode?
A:The projected light from the scanner is absorbed by the black lines and reflected by the white lines of the barcode, which creates a pattern on the optical sensor of the scanner.

B:The projected light from the scanner is refracted by the lines on the barcode, which changes the speed of the light hitting the optical sensor of the scanner.

C:The projected light from the scanner interferes with the reflected light from the barcode, and this combined beam of light is recognized by the optical sensor of the scanner.

D:The projected light from the scanner is diffracted by the barcode, which creates a pattern on the optical sensor of the scanner.
Chemistry
2 answers:
mina [271]3 years ago
6 0

Answer:

The Answer is A

Explanation:

Had it before

aliya0001 [1]3 years ago
4 0

Answer:

A

Explanation:

The barcode is being read by a laser that scans along the length of the sequence, reflecting more light from the white strips and less from the  black strips. Hope this helps!

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Explanation:

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An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose ass
KIM [24]

Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ_{1} - 1/λ_{2})

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*10^{-19} J.

h = 4.135*10^{-15} eVs

c = 3*10^{8} m/s

λ_{1} = 300 nm = 300*10^{-9} m

λ_{2} = 640 nm = 640*10^{-9} m

Thus:

ΔE = 4.135*10^{-15} eVs*3*10^{8} m/s*(\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m })

ΔE = 4.135*10^{-15}*3*10^{8}*1.77*10^{6} eV = 2.196 eV

6 0
3 years ago
What is the formula of the chromium(iii) complex that contains two ammonia and four thiocyanate (scn−) ligands?
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Answer is: formula of the complex is [Cr(NH₃)₂(SCN)₄<span>]</span>⁻<span>.
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3 years ago
Trisha and her lab partner filled a beaker with 150 milliliters of a 0.10 M HCl solution. How many moles of HCl are in the beake
oksian1 [2.3K]
M=mol/liter
We know that we have 150ml=.15 L and .1 mol of HCl
Rearranging the molarity equation, we get
mol=M*l
mol=(.15)(.1)
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7 0
3 years ago
Read 2 more answers
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