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Daniel [21]
2 years ago
6

Which part of a laser printer applies a positive charge to the paper that attracts the toner particles to it

Physics
1 answer:
Shalnov [3]2 years ago
6 0

The part of laser printer that applies a positive charge to the paper in order to attract the toner particles is known as transfer roller.

<h3 />

What is a laser printer:

A laser printer is a kind of printer that uses the electrostatic digital printing process to perform printing. It makes use of the static electricity and toner powder in place of liquid ink.

The toner is applied to specific areas which are dependent on the charge difference created or on the static electricity.

Following are the components of a laser printer:

  • Scanning unit:

        This unit of a laser printer generally consists of a laser diode, a

        scanning motor and a polygon mirror.
        It also consists of two-beam alignment lenses.

  • Cartridge unit:

        This unit of laser printer consists of three drums, namely primary

        charging roller (PCR), organic photoconductive drum (OPC) , and

        image transfer roller (ITR).
        The transfer roller is also present at a close vicinity of the  

        printer's  toner cartridge.

  • Fuser assembly unit:

        This unit of laser printer consists of a pressure roller and a fuser                roller, where the fuser roller assembly consists of a heating

        element.

Therefore, the transfer roller unit of a laser printer applies a positive charge to the paper that attracts the toner particles to it.

Learn more about laser printers here:

<u>brainly.com/question/5039703</u>

#SPJ4

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the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for
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3 0
1 year ago
ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of
sesenic [268]

Answer:

a) V_{2/1}=0.8m/s

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

V_{2/1} = V_{2} - V_{1} = 0.8m/s

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s

t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m  Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

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3 years ago
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it has to be 20 charecters just ignore this your answer is up there
7 0
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