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kari74 [83]
2 years 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)

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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.

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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]2 years ago
7 0

Answer:

pp

Explanation:

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You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel
Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

T = 2 \pi \sqrt{\frac{m}{k}}

squaring on both sides,

T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

y=T^2=0.0569x+0.0010

the hanging mass M is x here, so comparing the equation we know that

\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m

b) In order to find the mass of the spring we make similar process, so comparing,

\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g

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