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Marizza181 [45]
3 years ago
7

An electron moving in a wire collides again and again with atoms and travels an average distance between collisions that is call

ed the mean free path. If the mean free path is less in some metals, what can you say about the resistance of these metals? For a given conductor, what can you do to lengthen the mean free path?
Physics
1 answer:
hoa [83]3 years ago
4 0

Explanation:

A substance with a short , medium, free path has improved electron flow resistance and a higher electrical resistance . Heat applications impose more molecular chaos on all materials and shorten the track further, increasing resistance of most materials. So, just refresh the material to expand the course. In certain materials, when cooled to the minimum temperature, the conductivity is substantially increased.

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A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
2 years ago
The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti
KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

Let v is the velocity of the particle. Velocity of an object is given by :

v=\dfrac{dr(t)}{dt}

v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

v=(8i+4tj+6k)\ m/s

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

a=\dfrac{dv(t)}{dt}

a=\dfrac{d[8i+4tj+6k]}{dt}

a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

7 0
3 years ago
Use V=Voe^-t/RC and T=RC to derive T1/2=Tln2
Ksju [112]
Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.

3 0
3 years ago
What is the difference between parallel and series​
DiKsa [7]

In a series circuit, the sum of the voltages consumed by each individual resistance is equal to the source voltage. ... In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component.

5 0
4 years ago
One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d
stira [4]

Answer:

a) Diffusion  coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

^{2} = 2Dt..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

^{2} = 2Dt

0.1^{2}  = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0
3 years ago
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