Which of the following statements is true of electromagnetic waves but NOT mechanical waves.

HELP I WILL GIVE BRAINLYEST!!

zavuch27 [327]

Answer:

A-B-C

Explanation:

Depends if you have good taste.

8 0

3 years ago

A cube that is 20 nanometer on an edge contains 399,500 silicon atoms, and each silicon atom has 14 electrons and 14 protons. In

Sergeu [11.5K]

Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

N-type Impurities: Pentavalent impurities e.g. Phosphorous, Arsenic are added to have an additional electron in the structure. Thus a pentavalent impurity creates 1 additional electron.
P-type Impurities: Trivalent impurities e.g. Boron, Aluminium are added to have a positive "hole" in the structure. Thus a trivalent impurity creates 1 hole.

Now for estimation of extra electrons in the impured structure is as

$N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\$

Now for estimation of "holes" in the impured structure is as

$N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\$

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

$N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3$

So total 3 "holes" are available for conduction of current at 300K.

6 0

3 years ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago

Which way does wind blow

prisoha [69]

I think it blows vertically and horizontally cause wind can blow different directions

5 0

3 years ago

Read 2 more answers

Why did the waco tornado happen?

Natali5045456 [20]

it was too much water inside the clouds and the tornado happened

6 0

3 years ago