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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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It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.
Alenkasestr [34]
The work to stretch a spring from its rest position is

               (1/2) (spring constant) (distance of the stretch)²

           E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

                 1700 joules  =  1/2 k (3m)²

1 joule = 1 newton-meter

                 1700 N-m =  1/2 k  (3m)²

Multiply each side by 2:   3400 N-m  =  k · 9m²   

Divide each side by  9m²     k  =  3400 N-m / 9m²

                                                   =  (377 and 7/9)   newton per meter    

7 0
4 years ago
Read 2 more answers
A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r
bazaltina [42]

Answer:

V=16.65 m^3

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

R=\frac{25}{100}*74kg=18.5kg

So the net force uproad on the balloon is

F_b=18.5kg*g

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

P_h=\frac{m}{V}=0.179 kg/m^3

P_A=1.29 kg/m^3

F_b=F_A-F_H

F_b=m_a*g-m_h*g

m=P/V

18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g

V=\frac{18.5kg}{(1.29-0.179)kg/m^3}

V=16.65 m^3

4 0
3 years ago
Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo
bagirrra123 [75]
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span> =>v¹ = 2.25 m/s (East) </span>
5 0
3 years ago
A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastwar
tekilochka [14]
<h2>a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    \texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m

    Magnitude of his displacement = 2729.47 m

3 0
3 years ago
Read 2 more answers
1. What is the formula for the period of a pendulum and what is the main determining factor in its period?
Brut [27]

Answer:

T=2\pi \sqrt{\frac{L}{g}}

Explanation:

A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.

The period of a pendulum is the time it takes for the pendulum to complete one oscillation.

The period of a pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that the period of a pendulum does not depend on the mass.

Therefore, the only 2 factors affecting the period of a pendulum are:

- The length of the pendulum: the longer it is, the longer the period of oscillation

- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum

7 0
4 years ago
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