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Talja [164]
3 years ago
6

Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Elec

trons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are not quantized. When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. The energies of atoms are quantized.
Physics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer:

Explanation:

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>FALSE. </em>The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.

The energies of atoms are not quantized. <em>FALSE. </em>The energies of the atoms are in specific levels.

When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. <em>FALSE. </em>During absorption, a specific wavelength of light is absorbed, not emmited.

Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>TRUE. </em>Again, you can observe just the transition due the change of energy of an electron in the quantized energy level

When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. <em>TRUE. </em>The electron decreases its energy releasing a specific wavelength of light.

The energies of atoms are quantized. <em>TRUE. </em>In fact, the energy of all subatomic, atomic, and molecular particles is quantized.

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3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

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3 years ago
A car increases velocity from 25 m/s to 50 m/s in 5 seconds. What is the acceleration of the car?
Mars2501 [29]

Answer:

the acceleration of the car is 5m/s2

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Lourdes mixes several ingredients in a bowl, creating a cake batter. She holds the bowl up and turns it upside down, causing the
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friction and gravity ....................

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Two asteroids in outer space collide, and stick together. The mass of each asteroid, and the velocity of each asteroid before th
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Answer: (2) Use the Momentum Principle.

Explanation:

In fact, it is called the <u>Conservation of linear momentum principle,</u> which establishes the initial momentum p_{i} of the asteroids before the collision must be equal to the final momentum p_{f} after the collision, no matter if the collision was elastic or inelastic (in which the kinetic energy is not conserved).

In this sense, the linear momentum p of a body is defined as:

p=mV

Where m is the mass and V the velocity.

Therefore, the useful approach in this situation is<u> option (2)</u>.

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I dont know this but I have an app for you if your really strangling

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