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4 years ago

How much work is required to move

1 answer:

6 0

The work is path independent since we have a conservative force.

Thus $W=d\cdot\frac{q\cdot U}{d^2}=\frac{3.0\cdot9.0}{0.010}=\boxed{2700 J}$

Answer (1)

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Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

$P_x=rcos\theta$

$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

Velocity at y=150 mm

$150=300sin\theta$

thus $\theta =30^{\circ}$

position of $P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}$

$P_x=259.80 mm$

$P=259.80\hat{i}+150\hat{j}$

Velocity at this instant

$u_x=-rsin\theta =300\times sin30=-150 mm/s$

$u_y=rcos\theta =300\times cos30=259.80 mm/s$

4 0

3 years ago

I would like to know the answers to both 10 & 11

oksian1 [2.3K]

Answer:

picture is not showing up sorry

Explanation:

7 0

4 years ago

How many second does it take for sunlight to reach the surface of the earth it

sertanlavr [38]

Answer:

1 second

Explanation:

5 0

3 years ago

Read 2 more answers

Why does conduction occur in solids?

dmitriy555 [2]

Answer:

Conduction occurs more readily in solids and liquids, where the particles are closer to together, than in gases, where particles are further apart. ... As these molecules collide, thermal energy is transferred via conduction to the rest of the pan.

Explanation:

Metals have tightly packed atoms which can easily pass on their kinetic energy and also have free moving electrons.

5 0

3 years ago