What is the s. i unit of current

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.

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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:

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6 0

2 years ago

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

UkoKoshka [18]

Answer:

a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ = 1.6 mN (going in the positive x direction, towards q₂)

6 0

3 years ago

Read 2 more answers

Q6) A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadil

77julia77 [94]

Answer:

-2.8 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²

Using the equation of motion,

v² = u² + 2as................... Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,

Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.

Substituting into equation 1

6² = 8²+2(a)5

36 = 64 + 10a

10a = 36-64

10a = -28

10a/10 = -28/10

a = -2.8 m/s²

Note: a is negative because because the skater decelerate on the rough ice

Hence the magnitude of her acceleration is = -2.8 m/s²

6 0

3 years ago

A gas is placed in a storage tank at a pressure of 49.2 atm at 39.0C . As a safety device, a small metal plug in the tank is mad

Amiraneli [1.4K]

Answer:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=49.2 atm\\T_1=39.0^oC = 312.15 K\\P_2=?\\T_2=198^oC=471.15 K$

Putting values in above equation, we get:

$\frac{49.2atm }{312.15 K}=\frac{P_2}{471.15 K}\\\\P_2=74.26 atm$

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

4 0

3 years ago

A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure

klasskru [66]

Answer:

$3x_0$

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

$d=v_0 t = x_0$

in the second case, the horizontal velocity is increased to

$v_x' = 3v_0$

And so the new distance travelled will be

$d' = v_x' t = 3 v_0 t = 3 x_0$

So, the distance increases linearly with the horizontal velocity.

5 0

3 years ago

What is the s. i unit of current​

What is the s. i unit of current