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2 years ago

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a generator produces a maximum emf of 240 v when rotating at 3.7 rad/s in a magnetic field of 0.200 t. if the coil has 822 loops

what is the area of each loop?

1 answer:

Nesterboy [21]2 years ago

6 0

Answer:

I actually did that but I fprgot

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago

Which projectile will be visibly affected

jonny [76]

Answer:

leaf and a balloon is the correct answer

6 0

3 years ago

Read 2 more answers

If you sight Polaris with an astrolabe and determine its altitude is 20° above the horizon, then what does this tell you about y

oksano4ka [1.4K]

If you sight Polaris at 20 degrees above your Northern Horizon then you know that your latitude is 20 degrees north of the equator.

3 0

3 years ago

Read 2 more answers

i)Distinguish between different methods of charging. ii) You are provided with a positively charged gold leaf electroscope. Stat

AleksAgata [21]

Answer:

Explanation:

On rubbing a glass rod with silk, the electrons from the glass rod get transferred to the silk. The silk now has an excess of electrons and so is negatively-charged. On the other hand, the glass rod is deficient in electrons and hence is positively-charged.

In the above case, the silk undergoes negative electrification.

Now, when the positively charged glass rod is touched on the disc of a negatively charged gold leaf electroscope, the electrons shifts towards rod, hence amount of charge on gold leaves decreases and the divergence between the gold leaves decreases as unlike charges attract each other.

Hence, the divergence decreases when a glass rod rubbed with silk is brought near the disc of negatively charged electroscope.

hope it helps pls mark me as brainliest

6 0

3 years ago

Determine the number of atoms per unit cell in a (a) face-centered cubic, (b) body- centered cubic, and (c) diamond lattice.

fredd [130]

Answer:

a) 4

b) 2

c) 8

Explanation:

In a cubic lattice, each atom of the vertex is shared among other 8 unit cells, so the atoms on the vertex contribute to 1/8 to a given unit cell.

a) Ina face-centered cubic we have 8 atoms on the vertex and one in each face, which is shared with another unit cell, so it contribute to 1/2

Therefore, the total atoms are:

8*(1/8) + 6*(1/2) = 4

b) In the body centered cubic structure, the centered atom is not shared with another cell, therefore it contribute to 1 to the given cell:

The number of atoms per unit cel is:

8*(1/8) + 1 = 2

c) The diamond lattice is similar to the face-centered cubic lattice but it contains two identical atoms per lattice point.

Therefore it must contain twice atoms than the face-centered cubic lattice:

that is, it has 8 atoms per unit cell

3 0

3 years ago