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3 years ago

In an electronic transition an atom can not emit what?

1 answer:

7 0

█ Question <span>█

</span><span>In an electronic transition, an atom cannot emit what?

</span>█ Answer █

When an electronic transition is occurring, an atom cannot emit ultra-violet light.

<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>

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A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

When all else remains the same, what effect would decreasing the focal length have on a convex lens?

aniked [119]

<h3>Answer;</h3>

<u>It would make the lens stronger. </u>

<h3>Explanation;</h3>

The focal length is the distance between the optical center or the center of the lens to the focal point of a convex or concave lens.
The power of the convex lens is lens ability to undertake refraction or bend light. It is given as the reciprocal of focal length.
Power of the lens = 1/ f; therefore the smaller the focal length the higher the power and the larger the focal length the lower the power.
Thus; decreasing the focal length of a convex lens makes the lens stronger.

4 0

3 years ago

Read 2 more answers

What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high? (Formula: PE = mgh)

Lisa [10]

Answer:

The answer would be 735J

Explanation:

PE=mgh

=(mass)(force of gravity)(height)

=(25kg)(9.8m/s^2)(3m)

=735J

8 0

3 years ago

Read 2 more answers

A block of mass m = 2.0 kg slides head on into a spring of spring constant k = 260 N/m. When the block stops, it has compressed

nasty-shy [4]

Answer:

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Explanation:

a) The work done by the spring force

Ws = ½ * k * x²

Ws = ½ * 260 N/m *0.14² m

Ws = 2.548J

b) The increase in thermal energy can by find using

Et = Wf

Wf = µ * m *g * x

Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m

Wf = 1.153 J

c) The speed just as the block reaches can by fin using

EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s

8 0

3 years ago

What determines the voltage of a light bulb?

valentinak56 [21]

The power source its getting usually determines its voltage.

5 0

3 years ago