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GalinKa [24]
3 years ago
15

Explain the flow of electrons in p-type semiconductors and why it looks like a flow of positive charges.

Physics
2 answers:
Morgarella [4.7K]3 years ago
5 0

Answer:

In p type of semiconductors we know that impurity of trivalent valency is added into the pure semiconductors. Due to this trivalent impurity the bond is made between the atoms in such a way that it will show a vacant space in its bonding around an atom.

Due to the bonding of the trivalent valency of atom with the pure semiconductor atom it form a bond in which 3 electrons will share completely while in fourth valency it shows a vacancy.

This vacant space is known as holes. which will be filled by shared electrons as it is a lower energy state in the space.

Now when we see the charge sharing in this then here the vacant space is filled due to the shifting of valance electrons in the bonds. As electron will fill the position of hole then it will create the vacancy at the previous position of the electron and this continues till the vacancy is filled by different electron.

As we can see that holes are created at different positions due to flow of electrons then we can relate is to the flow of positive charge as its motion is opposite the flow of electrons.

Svetach [21]3 years ago
4 0
One of the key terms in describing the flow of electrons in p-type semiconductors is the concept of holes. Holes are generally produced when an electron is gone missing when its moves from a lower potential to a higher potential. What is left behind the process is a term called "hole."
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A roofer drops a nail that hits the ground traveling at 26 m/s. How fast was the nail traveling 1 second before it hits the grou
RoseWind [281]

vf = vi + at 

vf – vi = at<span>

<span>vi= 0, vf=26  and afor nil = 9.8m/s2</span></span>

26 = 9.8t

t =<span> 26 / 9.8 = 2.65 s
Now we know the total time, so we can calculate the time 1 second before it hit the ground. 

<span>= 2.65 -1 = 1.65s

<span>Now again using the same equation, vf = vi+at, we can find vf
vi = 0, a = 9.8 t=1.65</span></span></span>

vf = 0 + 9.8(1.65) = 16.17 m/s<span>
</span><span>So, the nail is traveling with the speed of 16.17m/s 1 second before it hits the ground.</span>

6 0
3 years ago
What metric units are used to report the mass of an object
sladkih [1.3K]
The unit used<span> to measure weight in the metric system is the gram.</span>
4 0
4 years ago
The Jamaican bobsled team was moving at a velocity of 50 m/s, then they hit the brakes on their sled to decelerate at a uniform
atroni [7]

Answer:

The time it took the bobsled to come to rest is 10 s.

Explanation:

Given;

initial velocity of the bobsled, u = 50 m/s

deceleration of the bobsled, a = - 5 m/s²

distance traveled, s = 250 m

Apply the following kinematic equation to determine the time of motion of the bobsled;

s = ut + ¹/₂at²

250 = 50t + ¹/₂(-5)t²

250 = 50t - ⁵/₂t²

500 = 100t - 5t²

100 = 20t -t²

t² - 20t + 100 = 0

t² -10t - 10t + 100 = 0

t (t - 10) - 10(t - 10) = 0

(t - 10)(t - 10) = 0

t = 10 s

Therefore, the time it took the bobsled to come to rest is 10 s.

3 0
3 years ago
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

g_e = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

g_e=G\frac{M}{r^2}

g=G\frac{0.9M}{r^2}

Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0
4 years ago
A cliff diver from the top of a 100 (m) cliff. He begins his dive by jumping up with a velocity of 5 (m/s) a. How
Nutka1998 [239]

Answer:

A.) 4.81 seconds

B.) 44.6 m/s

Explanation:

He begins his dive by jumping up with a velocity of 5 (m/s).

Let us first calculate the maximum height reached by using third equation of motion

V^2 = U^2 - 2gH

At maximum height, V = 0

0 = 5^5 - 2 × 9.8H

19.6H = 25

H = 25 /19.6

H = 1.28 m

The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.

Where height h = 1.28 + 100 = 101.28 m

h = Ut + 1/2gt^2

As the diver drop from maximum height, U = 0

101.28 = 1/2 × 9.8 × t^2

4.9t^2 = 101.28

t^2 = 101.28/4.9

t^2 = 20.669

t = sqrt ( 20.669)

t = 4.55s

As the diver jumped up, the time taken to reach the maximum height will be

Time = 1.28 / 5 = 0.256

The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds

B.) Velocity right before he hits the water will be

V^2 = U^2 + 2gH

But U = 0

V^2 = 2 × 9.8 × 101.28

V^2 = 1985.09

V = 44.6 m/s

6 0
4 years ago
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