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kirill115 [55]
2 years ago
6

(b) Show that for certain incident energies there is 100 percent transmission. Suppose that we model an atom as a one-dimensiona

l square well with a width of 1 A and that an electron with 0.7 eV of kinetic energy encounters the well. What must the depth of the well be for 100 percent transmission? This absence of scattering is observed when the target atoms are composed of noble gases such as krypton.
Physics
1 answer:
bixtya [17]2 years ago
7 0

Answer:

Explanation:

100 persent transmission implies that the T=1

Therefore using the previous result we have

1+\frac{\sin^2\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a}{4\frac{E}{V_0}\frac{(E+V_0)}{V_0}}=1


\sin\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a=0\Rightarrow \sqrt{\frac{2m}{\hbar^2}(E+V_0)}=0\Rightarrow E=-V_0


The depth of the well for 100% transmission should be

V_0=-0.7~{\rm{eV}}

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Answer:

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A person is pulling their 20 kg luggage using the luggage handle. The handle is at an angle of 56 degrees above the horizontal.
rusak2 [61]

Answer:

The answer to your question is:  a = 1.99 m/s²

Explanation:

Data

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angle = 56°

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horizontal acceleration = ?

Process

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                                           cos Ф = adjacent side / hypotenuse

                                          adjacent side = hypotenuse x cosФ

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                                  F = ma

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2 years ago
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Debora [2.8K]

Answer:

\mu_{k} \approx 0.719

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

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After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

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g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a

\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}

\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )

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5 0
3 years ago
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