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alex41 [277]
2 years ago
11

Does a photon emitted by a higher-wattage red light bulb have more energy than a photon emitted by a lower-wattage red bulb

Physics
1 answer:
Liono4ka [1.6K]2 years ago
6 0

Answer:

The energy of these two photons would be the same as long as their frequencies are the same (same color, assuming that the two bulbs emit at only one wavelength.)

Explanation:

The energy E of a photon is proportional to its frequency f. The constant of proportionality is Planck's Constant, h. This proportionality is known as the Planck-Einstein Relation.

E = h\, f.

The color of a beam of visible light depends on the frequency of the light. Assume that the two bulbs in this question each emits light of only one frequency (rather than a mix of light of different frequencies and colors.) Let f_{1} and f_{2} denote the frequency of the light from each bulb.

If the color of the red light from the two bulbs is the same, those two bulbs must emit light at the same frequency: f_{1} = f_{2}.

Thus, by the Planck-Einstein Relation, the energy of a photon from each bulb would also be the same:

h\, f_{1} = h\, f_{2}.

Note that among these two bulbs, the brighter one appears brighter soley because it emits more photons per unit area in unit time. While the energy of each photon stays the same, the bulb releases more energy by emitting more of these photons.

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Answer:

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2 years ago
A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
postnew [5]

Answer:

r=15.53 nm

F=9.57\times 10^{-13}N

Explanation:

Lets take electron is in between iron and uranium

Charge on electronq_1= -1.602\times 10^{-19}C

Charge on ironq_2= 2\times 1.602\times 10^{-19}C

Charge on uraniumq_3= 1.602\times 10^{-19}C

We know that force between two charge

F=K\dfrac{q_1 q_2}{r^2}  

K=9\times 10^9\dfrac{N-m^2}{c^2}

For equilibrium force between electron and iron should be force between electron and  uranium

Lets take distance between electron and  uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}  

K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}  

q_2= 2\timesq_1,q_3=q_1

\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}

So r=15.53 nm

So force

F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}  

F=9.57\times 10^{-13}N

7 0
2 years ago
Suppose you are in total equilibrium in water (in other words you are floating and not moving in any direction)What could you do
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Explanation:

According to newtons first law of motion:

'' a body will continue in its state of rest or uniform motion along a path unless it is acted upon by an external force".

A body in equilibrium that is floating will be stable and not move in any direction. Even if it moves, the motion will be constant wouldn't change.

  • To move the body in any direction, one has to swim.
  • Swimming is the application of an external force to counter the balanced forces at equilibrium on a body.
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learn more:

Newton brainly.com/question/11411375

#learnwithBrainly

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