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astra-53 [7]
3 years ago
5

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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The formation of new volcanoes and the eruption of some existing ones already. 
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A bat hits a 0.150 kg baseball for
lara31 [8.8K]

Answer:

11.85 kg m/s

Explanation:

impulse = mass ( change in velocity )

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               = 0.150 ( 32.0 - (-47.0))

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3 0
4 years ago
The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a
dimaraw [331]

The absolute pressure is given by the equation,

P_{abs}=P_{atm}-P_{vac}

Here, P_{abs} is absolute pressure,P_{atm} is atmospheric pressure and P_{vac} is vacuum pressure.

Therefore,

P_{abs}=98 kPa-80 kPa=18kPa

The gage pressure is given by the equation,

P_{gage}=P_{abs}-P_{atm}.

Thus,

P_{gage}=18kPa-98 kPa=-80 kPa.

In kn/m^2,

The absolute pressure,

P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2

The gage pressure,

P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2.

In lbf/in2

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2

The gage pressure,

P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2

In psi,

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi.

The gage pressure,

P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi

In  mm Hg

The absolute pressure,

P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg

The gage pressure,

P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg



3 0
3 years ago
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo
Helga [31]

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

F_s=ma_{min}

\mu_sN=ma_{min}

\mu_s\cdot mg=ma_{min}

a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2

6 0
3 years ago
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the
pantera1 [17]

Answer:

Explanation:

F = k*n

F = 30N

k = ???

N = 0.73 M

F = k* N

30N = k * 0.73

k = 30N / 0.73

k = 41.1

5 0
3 years ago
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