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Lina20 [59]
1 year ago
13

ONLY ANSWER IF YOU KNOW 100% BRAINLIEST TO THE FIRST CORRECT ANSWER, AND ANSWER ASAP!

Physics
2 answers:
Vinvika [58]1 year ago
7 0
A i just did it on edg
saul85 [17]1 year ago
5 0

Answer: Meter and second

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Which part of a laser printer applies a positive charge to the paper that attracts the toner particles to it
Shalnov [3]

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

6 0
2 years ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

5 0
3 years ago
A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli
Andrews [41]

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

4 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
4 years ago
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