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nikklg [1K]
3 years ago
6

collection of such objects we observe that they emit electromagnetic radiation of three different energies: 0.7 eV (infrared), 2

.1 eV (visible), and 2.8 eV (visible). (a) Draw a possible energy-level diagram for one of the quantum objects, which has three bound states. On the diagram, indicate the transitions corresponding to the emitted photons, and check that the possible transitions produce the observed photons and no others. When you are sure that your energy-level diagram is consistent with the observed photon energies, enter the energies of each level (K+U, which is negative). Enter ALL levels before submitting; all of the energies must be correct to be properly scored. The energy K+U of the ground state is -5 eV. eV = energy of highest level (2nd excited state)
Physics
1 answer:
UkoKoshka [18]3 years ago
4 0

It 45 because it has to be this and that and ur gay

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The spiral review which of the following represents 7 + 7 + 7+7
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Do I have to choose physics to be an architect
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7 0
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What three factors affect the acceleration of an object?
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Explanation:

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7 0
3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
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