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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.

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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]1 year ago
7 0

Answer:

pp

Explanation:

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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
2 years ago
Please help! will give brainlest!!!!!!!!!!!!
eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

a=0

Therefore the net force is zero as well:

\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

where

F cos \theta is the horizontal component of the applied force, with

F = 22.5 N

\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0
2 years ago
A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the
topjm [15]
<h2>5.3 km</h2>

Explanation:

       This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

       Let us denote each of the individual displacements by a vector. Consider the unit vectors \vec{i}\textrm{ and }\vec{j} as the unit vectors in the direction of East and North respectively.

       By simple calculations, we can derive the unit vectors \vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2} in the directions North, 45^{o} South of West and 60^{o} North of West respectively.

       So Total displacement vector = Sum of individual displacement vectors.

       Displacement vector = 4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}

       Magnitude of Displacement = |-4.25\vec{i}+3.165\vec{j}|=5.3km

∴ Total displacement = 5.3km

4 0
2 years ago
When a 440-Hz tuning fork and a piano key are struck together, five beats are heard. If the pitch of the note on the piano is lo
vovangra [49]
The frequency would also be lower
7 0
2 years ago
Read 2 more answers
Can someone help with my physics homework? please
Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0
2 years ago
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