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3 years ago

Consider a one-dimensional crystal (similar to a carbon nanowire) with length 10 um and lattice spacing 0.1 nm.

Physics

1 answer:

Finger [1]3 years ago

5 0

Answer:

a) Fermi level = 600 electron-volts

b) $\frac{2.04 * 10^{13} }{\sqrt{E} }$

Explanation:

Given data:

length of one-dimensional crystal = 10 um

Lattice spacing = 0.1 nm

A) Determine the Fermi level assuming one electron per atom

Total length = 10 um

Interatomic separation of a = 0.1 nm

in this case the Atom has one electron therefore the number of electrons = 10^5 and the number of states Ns = gsN = 2 * 10^5 ( attached below is some part of the solution )

hence : Fermi level = 600 electron-volts

B) Determine the density of states as a function of electron energy

attached below is the detailed solution

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A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient

e-lub [12.9K]

Answer:

The coefficient of static friction between the crate and the floor is 0.41

Explanation:

Friction Force

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

$Fr=\mu N$ [1]

Where $\mu$ is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for $\mu$ :

$\displaystyle \mu=\frac{Fr}{N}$

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

$\displaystyle \mu=\frac{80}{196}$

$\mu=0.41$

The coefficient of static friction between the crate and the floor is 0.41

7 0

3 years ago

A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s

Andrej [43]

Answer:

$1300 m$

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from $20 s$ to $70 s$ is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area $=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)$

$=200+1000+100$

$=1300$

As velocity is in $m/s$ and time is in $s$ so the unit of area is $m$

Hence, the total distance is $1300m$ .

8 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$