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kari74 [83]
1 year ago
10

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o

f length L and cross-sectional area A lies along the x-axis between x=0 and x=L. The material obeys Ohm's law, and its resistivity varies along the rod according to ?(x)=?0exp(?x/L). The end of the rod at x=0 is at a potential V0 greater than the end at x=L.
Find the total resistance of the rod.

Express your answer in terms of the given quantities and appropriate constants.

R =
.632L?0A

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Correct

Part B

Find the current in the rod.

Express your answer in terms of the given quantities and appropriate constants.

I =
V0A?0L(1?e?1)

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Correct

Part C

Find the electric-field magnitude E(x) in the rod as a function of x.

Express your answer in terms of the given quantities and appropriate constants.

E(x) =
V0e?xLL(1?e?1)

SubmitMy AnswersGive Up

Correct

Part D

Find the electric potential V(x) in the rod as a function of x.

Express your answer in terms of the given quantities and appropriate constants.

V(x) =
V0e?xL?e?11?e?1

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Correct

Part E

Graph the function ?(x) for values of x between x=0 and x=L.

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Incorrect; Try Again; 2 attempts remaining

Part F

Graph the function E(x) for values of x between x=0 and x=L.

SubmitMy AnswersGive Up

Incorrect; Try Again; 4 attempts remaining

Part G

Graph the function V0(x) for values of x between x=0 and x=L.
Physics
1 answer:
zavuch27 [327]1 year ago
7 0

Answer:

pp

Explanation:

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A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 C. What would its temperature be (in C) when its pressure is increa
Harlamova29_29 [7]

Answer:

T₂ = 111.57 °C

Explanation:

Given that

Initial pressure P₁ = 9.8 atm

T₁ = 32°C  = 273 + 32 =305  K

The final pressure   P₂ = 11.2 atm

Lets take the final temperature = T₂

We know that ,the ideal gas equation  

If the volume  of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

T_2=\dfrac{P_2}{P_1}\times T_1

Now by putting the values in the above equation ,we get

T_2=\dfrac{11.2}{9.8}\times 305\ K

T_2=348.57\ K

T₂ = 384.57 - 273 °C

T₂ = 111.57 °C

3 0
2 years ago
A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b
Degger [83]

Answer:

The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

Explanation:

Given that,

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We need to find the correct relationships

Using balance equation

In horizontally,

The acceleration is a

T-Fd=ma...(I)

In vertically,

No acceleration

w=L

mg-L=0

Put the value of mg

L-fg=0....(II)

Hence,  The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

3 0
2 years ago
A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit
Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= -  GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}

=  - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}

=   - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0
1 year ago
A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine
Alex Ar [27]

Given


m1(mass of red bumper): 225 Kg


m2 (mass of blue bumper): 180 Kg


m3(mass of green bumper):150 Kg


v1 (velocity of red bumper): 3.0 m/s


v2 (final velocity of the combined bumpers): ?




The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:


Pa= Pb


Where Pa is the momentum before collision and Pb is the momentum after collision.


Now applying this law for the above problem we get


Momentum before collision= momentum after collision.


Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s


Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

Now we know that Momentum before collision= momentum after collision.


Hence we get


1215 = 555 v2


v2 = 2.188 m/s


Hence the velocity of the combined bumper cars is 2.188 m/s

4 0
1 year ago
Read 2 more answers
During a family trip to Laura's grandmother's house, the family cast traveled a distance of of 8 miles in 24 minutes. During the
Vanyuwa [196]

The correct answer is B

4 0
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