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Dmitrij [34]
3 years ago
5

On arrival at a vehicle collision, you observe a small fire in the engine compartment. A bystander is attempting to smother the

fire with a jacket as you approach with an A-B-C extinguisher from your ambulance. Which of the following is the best method of extinguishing the fire?
a. Sweep the nozzle of the extinguisher left and right using short bursts until you are out of extinguishing agent.
b. Aim at the base of the fire and use short bursts until the fire is out.
c. Close the hood to contain the fire within the engine compartment.
d. Aim the nozzle to the side of the fire and, with a sweeping motion, push the fire at its base
Physics
1 answer:
mina [271]3 years ago
3 0

Answer:

<em>Aim at the base of the fire and use short bursts until the fire is out.</em>

<em></em>

Explanation:

Fire extinguishers use CO2 (Carbondioxide) as the extinguishing agent. This is because CO2 is denser than air, and does not support combustion.

Aiming at the base of the fire causes the CO2 to fall on the base of the fire, where the source of the fire is, trapping it, and preventing it from further reacting with air in a combustion reaction. Also, the short burst creates a strong wind that forces the flame to blow out.

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3 years ago
In the Zeeman effect, the energy levels of hydrogen are split by a magnetic field. Each state with a different value of mlml has
Nikolay [14]

Answer:

the Zeeman effect without spin is three lines

Explanation:

The Zeeman effect is the result of the interaction of the magnetic field with the orbital angular momentum of the electrons, if we do not take the spin into account it is called the Normal Zeeman effect.

Therefore we only take into account the orbital moments (m_l) of the transition, from the selection rules of the refreshed harmonics, only the transition with

             \Delta m_l = 0, ± 1

            ΔE \DeltaE = \mu_B \ \Delta  m_l \ B

Let's analyze for the case of the Hydrogen atom

For a transition between levels n = 1 and n = 2 the values ​​of m_l are n_f = 1 m_l = 0 and for n₀ = 2  m_l = 0, 1

so we only have two lines.

For transition n_f = 2 and n₀ = 3

n_f = 2    m_l = 0, 1

n₀ = 3      m_l = -1, 0, 1

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for n_f = 3 and n_o = 4

n_f = 3   ml = -1, 0, 1

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Only transitions with  Δm_l = ±1 are allowed, so there are only two transitions plus the central transition (Δm_l = 0), so there are only 3 spectral lines.

In summary, due to the selection rule of spherical harmonics, the greatest number of lines in the Zeeman effect without spin is three lines.

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