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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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A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 2.79 m/s2, directed downward. (a) What
DedPeter [7]

Answer:

The vertical force acting on the firefighter = 908.27 N

Explanation:

Force: Force of a body is defined as the product of mass and its acceleration. The S.I unit of force is Newton (N)

The vertical force acting on the firefighter = Force due to the weight of the firefighter + force due to acceleration.

Ft = Fw - Fa

Where Ft = The vertical force acting on the firefighter, Fw = Force due to the weight of the firefighter, Fa = force due to acceleration.

Fw = mg

Making m the subject of formula in the equation above

m = Fw/g................... Equation 1

Where m = mass of the firefighter, g = acceleration due to gravity,

<em>Given: Fw = 707 N, </em>

<em> Constant: g = 9.8 m/s²</em>

Substituting these values into eqaution 1

m = 707/9.8

m = 72.14 kg.

But, Fa = ma

Where a = acceleration of the firefighter.

<em>Given: a = 2.79 m/s², </em>

<em>And m = 72.14 kg</em>

Fa = 72.14 × 2.79

Fa = 201.27 N

Therefore, Ft = 707 + 201.27  = 908.27 N

Ft = 908.27 N

The vertical force acting on the firefighter = 908.27 N

7 0
3 years ago
A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
2 years ago
2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA
11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

5 0
3 years ago
In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

8 0
3 years ago
The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso
nalin [4]

Explanation:

It s given that,

Mass of a planet, M=3.7\times 10^{24}\ kg

Radius of a planet, R=9.2\times 10^{6}\ m

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

g=\dfrac{GM}{R^2}

g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}

g=2.91\ m/s^2

(2) The escape velocity is given by :

v=\sqrt{\dfrac{2GM}{R}}

v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}

v = 7324.61 m/s

Hence, this is the required solution.

3 0
3 years ago
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