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Alenkasestr [34]
3 years ago
13

The resistivity of a certain semi-metal is 10-3 Ohm-cm. Suppose we would like to prepare a silicon wafer with the same resistivi

ty. Assuming we will use n-type dopants only, what dopant density would we choose

Physics
1 answer:
faltersainse [42]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

  The dopant density is ND   ≈  8.135*10¹²  cm⁻³

Explanation:

The explanation is shown on the second , third ,fourth and fifth uploaded image

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3 0
4 years ago
A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum
uysha [10]

To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

C = \frac{Q}{V}

Here,

C = Capacitance

V = Potential difference between the plates

Q = Charge between the capacitor plates

At the same time the energy stored in the capacitor can be defined as,

U = \frac{1}{2} CV^2

We will start by finding the value of the capacitance, so we will have to,

C = \frac{2\mu C}{12.0V}

C = 0.166\mu F

Finally using the expression for the energy we have that,

U = \frac{1}{2} CV^2

U = \frac{1}{2} (0.166\muF)(12.0V)^2

U = 11.0*10^{-6} J

Therefore the minimum amount of work that must be done in charging this capacitor is 11.0*10^{-6} J

8 0
4 years ago
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A train travels 94 kilometers in 5 hours, and then 84 kilometers in 3 hours. What is its average speed?
diamong [38]
Average speed equal to = ,total distance traveled/total time taken
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3 0
3 years ago
A baseball player slides into third base with an initial speed of 4.0 m/s . if the coefficient of kinetic friction between the p
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The work done by the friction force to stop the player is equal to his loss of kinetic energy:
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\Delta K=K_i =  \frac{1}{2}mv^2

Substituting into the first equation, we get:
\mu m g d=  \frac{1}{2}mv^2
from which we find d, the distance covered by the player:
d= \frac{v^2}{2 \mu g}= \frac{(4.0 m/s)^2}{2 \cdot 0.4 \cdot 9.81 m/s^2}=2.04 m
4 0
3 years ago
A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from
kogti [31]

Answer:

32

Explanation:

7 0
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