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Alex73 [517]
2 years ago
12

Prob. 3: Manifestation of quantum phenomena (total 25 points) (a)-(e) 5 points each.

Physics
1 answer:
TEA [102]2 years ago
6 0

Answer:

3A. This phenomenon can be seen in the discrete emission of the molecules.

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams

3D. This is due to the stimulated emission

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)

Explanation:

This problem asks for some experimental explanations of various quantum phenomena.

3A. This phenomenon can be seen in the discrete emission of the molecules.

In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed

In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.

3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .

This is widely used for holographic and interference work

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.

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jenyasd209 [6]

Answer:

4 m/s^2

Explanation:

The acceleration is defined as: Δv/Δt (the difference of the velocity over a time period in which happens that difference).

Remember that a difference is calculated by subtracting the initial value of a physical quantity from its final value.

In our case:

Δv = Vfinal - Vinitial = 36m/s - 0 m/s = 36m/s

Δt = 9s

a = Δv/Δt = 36m/s / 9s = 4m/s^2

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Answer:

the rope should break

Explanation:

she with equal amounts of pulling are on each side then the rope should slowly start to tare apart and snap/break.

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Based on its orbit, which planet behaves the least like the others?
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2 years ago
A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

8 0
2 years ago
A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is
dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

  • <em>Mass of the pitcher, m₁ = 50 kg</em>
  • <em>Mass of the baseball, m₂ = 0.15 kg</em>
  • <em>Velocity of the ball, u₂ = 35 m/s</em>

<em />

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

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