Which mineral is also an element graphite courts calcite hematite

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

S.I unit for moment of inertia of a fly wheel

qaws [65]

Answer:

m is expressed in kilograms and r in metres, with I (moment of inertia) having the dimension kilogram-metre square.

8 0

1 year ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

Two identical positive charges are placed near each other. At the point halfway between the two chargesTwo identical positive ch

Nitella [24]

Answer:The electric field is zero and the potential is positive.

Explanation:

Two identical positive charges are separated by a certain distance and midway between charges two identical positive charges are placed near each other.

So the Electric field at midway is zero because the electric field due to both charges add up to give zero electric field.(because they point in opposite direction)

Potential is scalar quantity and charges are positive so they add up to give potential.

7 0

3 years ago

2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in

aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

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6 0

1 year ago