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kodGreya [7K]
4 years ago
8

A square conducting plate 52.0 cm on a side and with no net charge is placed in a region, where there is a uniform electric fiel

d of 82.0 kN/C directed to the right and perpendicular to the plate.
Find
(a) the charge density of each face of the plate and
(b) the total charge on each face.
Physics
1 answer:
Mrac [35]4 years ago
5 0

Explanation:

Given:  uniform electric field E= 82.0 kN/C.

a) charge density σ =ε_0 E.

therefore, \sigma =82\times10^3\times3.85\times10^{-12}\\=0.0000003157= 315.7 nC/m^2

b)Total charge on each face = σA

q=σA

=315.7\times10^{-9}\times52\times10^{-4}\\=1.614\times10^{-9}= 1.614 \text{ nC}\\\text{Similarly on the other face }  = -1.614 \text{ nC}

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A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi
EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0
2 years ago
A 330 kg piano slides 3.6 m down a 28o incline and is kept from accelerating by a man who is pushing back on it parallel to the
e-lub [12.9K]

Answer:

a. 652.68N

b. -2349.65J

c. -3116.12J

d. 5465.77J

e. Zero

Explanation:

a. According to equilibrium of forces, the force of gravity is equal to the sum of the frictional force and force exerted by the man in the opposite direction (since they're both resistant forces).

Fg = Fm + Fr

Fm = Fg - Fr

Fm = mgsin(28°) - umgcos(28°)

u = coefficient of frictional force.

Fm = 330*9.8*sin28 - 0.4*330*9.8*cos28

Fm = 1518.27 - 865.59

Fm = 652.68N

b. Work done by man is:

Wm = -Fm * d

Wm = -652.68 * 3.6

Wm = -2349.65J

c. Work done by friction force:

W(Fr) = -Fr * d

W(Fr) = -865.59 * 3.6

W(Fr) = -3116.12J

d. Work done by gravity:

Wg = Fg * d

Wg = 1518.27 * 3. 6

Wg = 5465.77J

e. Net work done on the piano is:

Work done by friction + work done by gravity + work done by man

= -3116.12 + 5464.77 + (-2349.65)

= 0J

7 0
3 years ago
Diamond shine brightly but the pieces of glass don't​
icang [17]

Answer:

thatss deeeeepppp

Explanation:

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5 0
3 years ago
HELP PLEASE!!
sweet-ann [11.9K]

Answer:

C. More of the heat is transferred to the kinetic energy of the copper atoms than to the kinetic energy of the water molecules.

Explanation:

Both equal masses of water and copper were heated at the same temperature. Since copper is a good conductor of heat compared to water, its  absorbs more heat. Which in-turn increases the rate of vibrations of the atoms in the copper mass, thus increasing their kinetic energy.

In the case of water, its molecules displaces one another after being heated to a higher temperature compared to neighboring molecules. So that the heated molecule becomes less dense and floats to the surface of water.

This property of copper makes it to be heated to a higher final temperature than the water.

6 0
3 years ago
Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

Answer:

\Delta U = 1640 J

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

W = 560 J

so we have

\Delta U = Q - W

\Delta U = 2200 - 560

\Delta U = 1640 J

6 0
4 years ago
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