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alex41 [277]
3 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]3 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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You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential
Dahasolnce [82]
True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...
3 0
2 years ago
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
If two or more elements are involved in a chemical reaction they will most likely form a?
Vitek1552 [10]
I know either mixture or compound

5 0
3 years ago
Read 2 more answers
A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci
vovikov84 [41]
According to the law of conservation of momentum:

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know. 

m1 = 125 kg     v1 = 12m/s      v'1 = -12.5m/s
m2 = 235kg      v2 = -13m/s     v'2 = ?

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'
(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'
1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')
-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation. 

-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')
7.5kg.m/s=(235kg)(v_{2}')
(7.5kg.m/s)/(235kg)=(v_{2}')
0.03m/s=(v_{2}')

The velocity of the 2nd car after the collision is 0.03m/s.
5 0
3 years ago
"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree
kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound.  It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.
 
6 0
3 years ago
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