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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 <em>u</em>m

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 bucket filled with water has a mass of 70kg and is hanging from a rope that is wound around a 0.054 m radius stationary cylind

guajiro [1.7K]

Answer:

T = 37.08 [N*m]

Explanation:

We must remember that torque is defined as the product of a force by a distance. This distance is measured from the point of application of force to the center of rotation of the rotating body.

The force is equal to the product of mass by gravitational acceleration.

$F=m*g\\F=70*9.81\\F=686.7[N]$

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$T=F*r\\T=686.7*0.054\\T=37.08[N*m]$

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3 years ago

Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o

natulia [17]

Answer

given,

Speed of car A = 95 Km/h

= 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

= 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

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3 years ago

on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak

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Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>

That's about 2.9 G's . Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't. Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds ! Maybe the part of the airplane that the catapult pushes on can't handle any more force than that. Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds. Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes ! Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

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3 years ago

If objects at the same distance suddenly decreased in mass the gravitational force between them would

inn [45]

Answer:

The Gravitational force would decrease

Explanation:

To calculate FG the formula is:

Fg=mg

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3 years ago