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kari74 [83]
3 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]3 years ago
7 0

Answer:

pp

Explanation:

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When a pitcher throws a baseball, it reaches a top speed of 39 m/s. if the
Aliun [14]

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Given that;

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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
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