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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.

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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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A wheel is rotating freely at angular speed 660 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi
sweet-ann [11.9K]

Answer:

(a)  110 rev/ min

(b) 5/6

Explanation:

As per the conservation of linear momentum,

L ( initial ) = L ( final )

I' ω' = ( I' + I'' ) ωf

I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.

(a)

So,    ωf = I' ω' /  ( I' + I'' )

As I'' = 5I'

ωf = I' ω' /  ( I' + 5I' )

ωf = ω'/ 6

now we know ω' = 660 rev /  min

therefore    ωf = 660/6

                       = 110 rev/ min

(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be   K'' =  ( I' + I'' )ωf² / 2

                                   K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

                                   K'' = 6I' ω'²/72

                                   K'' = I' ω'²/ 12

therefore the fraction lost is

                ΔK/K' = ( K' - K'' ) / K'

                           =  {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

                            = 5/6

6 0
3 years ago
Particles q1 = +8.0 UC, 92 = +3.5 uc, and
Olin [163]

Answer:

   F_total = 29.4 N,    directed to the right of particle 2

Explanation:

We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.

Let's use Coulomb's law to calculate each force

         F = k \frac{q_1q_2}{r_{12}^2}

particles 1 and 2

q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m

         F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²

         F₁₂ = 2.59 10¹ N

Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.

particles 2 and 3

q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m

we calculate

        F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²

        F₂₃ = 3.5 N

as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2

Now we add the forces as vectors

        F_total = ∑ F = F₁₂ + F₂₃

        F_total = 25.2 +3.5

        F_total = 29.4 N

directed to the right of particle 2

5 0
2 years ago
Read 2 more answers
Can someone please answer this, ill give you brainliest and your getting 100 points.
dsp73
It’s c hopefully that helps
5 0
2 years ago
Read 2 more answers
After an earthquake, which type of seismic wave arrives last at a seismometer?
amm1812
I believe L waves arrive last at a seismometer.
hope this helps!
8 0
2 years ago
The driver of a car traveling at a speed of 27 m/s slams on the brakes and comes to a stop in 3 s. If we assume that the speed c
NISA [10]

Answer:

Average speed = 13.5 m/s

Explanation:

Since the car is running at a speed of 27 m/s  and it stops after 3 seconds by applying the brake. Therefore, the initial speed of the 27 m/s and final speed is 0.

Use below formula to find the average speed :

Average speed = (Initial speed + final speed ) / 2  

Average speed = (27 + 0 ) / 2

Average speed = 13.5 m/s

7 0
3 years ago
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