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andrey2020 [161]
3 years ago
13

A material kept at high temperature is seen to emit photons with energies of 0.3 eV, 0.5 eV, 0.8 eV, 2.0 eV, 2.5 eV, and 2.8 eV.

These are the only photon energies observed. It is now cooled down to a very low temperature so that it is not emitting photons anymore. If a beam of light with a continuous range of energies from 0.01 eV to 10 eV shines on the material, what photon energies in this beam will correspond to dark absorption lines?
Physics
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

0.3 eV, 0.5eV,, 8 eV, 2.0eV, 2.50 eV, 2.8 eV

Explanation:

In a given material the emission and absorption spectra are equivalent, for which the emission spectrum observed at high temperature for the material corresponds to the transition between the energy states of the material, the process is that the electrons exist from the ground state until an excited state and after a short period of time or these electrons relax emitting photons.

In the absorption process, the material is at low temperature, ideally at A = 0K, whereby all states are in the ground state and all excited states are empty. therefore it can absorb the beam energy for each transition given from the ground state to each excited edtado.

Consequently, the lines above the absorption oscillate lines coincide with the lines of emotion, this we see lines oscillate at 0.3 eV, 0.5eV,, 8 eV, 2.0eV, 2.50 eV, 2.8 eV

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9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?

6 0
3 years ago
It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne
sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x  10⁷ / ω²

=2 x  10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

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R= 3.683 m .

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3 0
3 years ago
Why is the overall charge of the atom neutral or zero?
MrRissso [65]

Answer:

B

Explanation:

this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.

and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer

if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral

let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.

instead, write, the it is neutral

hope it helps if not please report it so that someone else gets to try it out

7 0
3 years ago
A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri
Drupady [299]

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

KE = I/2*\omega_o^2

The inertia is:

I=m/2*R^2=0.072kg.m^2

So, the kinetic energy will be:

KE = 888.26J

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf    where Kf=0;   Ko = 888.26J;  and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R   where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

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6 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel
Zanzabum
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
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Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
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Answer: ymax = 10084.2m
8 0
3 years ago
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