Explain why the wave model of light cannot explain the energy emissions from a blackbody
radiator, but the particle model of light can.
HELP MEEE ASAP

Answers

Answer 1

Answer:

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.

So as you see the wavelengths are in the x axis so all wavelengths are covered.

Black-body radiation provides insight into the thermodynamic equilibrium state of cavity radiation. If each Fourier mode of the equilibrium radiation in an otherwise empty cavity with perfectly reflective walls is considered as a degree of freedom capable of exchanging energy, then, according to the equipartition theorem of classical physics, there would be an equal amount of energy in each mode. Since there are an infinite number of modes this implies infinite heat capacity (infinite energy at any non-zero temperature), as well as an unphysical spectrum of emitted radiation that grows without bound with increasing frequency, a problem known as the ultraviolet catastrophe. Instead, in quantum theory the occupation numbers of the modes are quantized, cutting off the spectrum at high frequency in agreement with experimental observation and resolving the catastrophe. The study of the laws of black bodies and the failure of classical physics to describe them helped establish the foundations of quantum mechanics.

The above explains why the classical assumptions lead to a wrong spectrum.

Explanation:

i don't know if It helps you..parang Ang layo naman Ng sagot ko sa tanong mo

Answer 2

The wave model of light cannot explain the energy emissions from a blackbody radiator, but the particle model of light can be because the electromagnetic wave theory does not explain the black body radiation and the particle model of light can explain it.

What is Black body radiation?

Black body radiation is defined as the surface which absorbs all the energy and radiant light falling on it because it absorbs all light of color.

An example of black body radiation is a cavity with a small hole in it.

What is an Electromagnetic Wave ?

The Electromagnetic Wave is also called EM Waves which is defined as the waves that are created as a vibration between a magnetic field and electric field.

What is Light ?

Light is defined as the electromagnetic radiation which propogates as waves and allows us to make the object visible.

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Related Questions

A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.

Answers

Point A would be the build up and leverage to slide down the half pipe to point b she would be bending her knees to increase speed to go up the other side and meet point c

What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal

Answers

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

Therefore, the electric field is  3.91 x 10⁶ N/C

can someone please help me ​

Answers

F = 153 N

[tex]\theta = 11.3°[/tex]

Explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:

[tex]F_{x}[/tex] = 350 N - 200 N = 150 N

[tex]F_{y}[/tex] = 180 N - 150 N = 30 N

The magnitude of the resultant force F is given by

[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]

[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]

[tex]\:\:\:\:\:\:=153\:N[/tex]

To find the direction [tex]\theta[/tex], we use

[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]

or

[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]

A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 830 kg car to rest from an initial speed of 1.4 m/s.

Answers

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =[tex]F^>[/tex] × [tex]x^>[/tex]

W = [tex]Fxcos\theta[/tex]    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = [tex]Fxcos([/tex] 0° [tex])[/tex]

W = [tex]Fx[/tex] ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

[tex]W_{net[/tex] = ΔKE

= [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu²

we make F, the subject of formula

F = [tex]\frac{m}{2x}[/tex]( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

8 points)Antireflection coating can be used on the eyeglasses to reduce the reflection of light: a) A 100nm thick coating is applied to the lens. What must be the coating’s index of refraction to be most effective at 500nm? (Assume the coating index of refraction is less than that of the lens). b) If the index of refraction of the coating is 1.20, find the necessary thickness of the coating at 500nm.

Answers

Answer:

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

thickness of coating t = ?

t = λ / 4n

we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm

. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content of the system.

Answers

Answer:

40479.6 J

Explanation:

Applying,

q = cm(t₂-t₁).................... Equation 1

Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C

Constant: c = 4200 J/kg.°C

Substitute these values into equation 1

q = 4200(0.079)(143-21)

q = 331.8(122)

q = 40479.6 J

A bucket of mass m in a well is held up by a rope. the rope is wound around a drum of radius r there is also a handle of length R attached to the drum. the tension in the rope is equal to T. If the buket is allowed to fall into the well, which point will have the greatest angular acceleration, a point on the rim of the drum (at radius r) or a point on the end of the handle (at radius R)?

a. The point on the rim of the drum.
b. The point at the end of the handle.
c. They will both have the same angular acceleration.

Answers

Answer:

the correct answer is C

Explanation:

This is a system with circular motion, there is a relationship between the linear and angular variables

         a = α r

with the cube going down the well, the tension of the leather is maintained therefore the acceleration of the cube is

          W = m a

          -mg = ma

          a = -g

this acceleration a is the same as that at the edge of the drum.

         α = a / r

where we can see that the angular acceleration is constant

consequently the correct answer is C

Question 2:
Inclined Plane
A block (M) weighs 25-N, rests on an inclined plane when it is joined by a sting to a support
(S) as shown in the figure' below. Use g=10 N/Kg.
(S)
B
M
List and classify the forces acting on (M).
Représent, without scaling, the forces acting on (M).
Find the mass of (M).
74. If the string were cut, (M) does not slide. Explain this phenomenon.
15. Determine the mass and weight of (M) on moon.
06​

Answers

Answer:

we need the block

Explanation:

1×2 =4 lest 74 =345

A conducting sphere of radius R carries an excess positive charge and is very far from any other charges. Draw the graphs that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere?

Answers

Answer:

See annex

Explanation:

By convention potential at ∞    V(∞ ) = 0

As the distance from the sphere decreases the potential increases up to the point d = R  ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.

I am sorry I could not make a better graph

The graph that best  illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below

[tex]V = \frac{KQ}{R}[/tex]

for r <= R

[tex]V = \frac{KQ}{r}[/tex]

for  r > R  

Therefore the graph will be

For more information on potentials as function of distance

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The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.​

Answers

Explanation:

Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.

Nick and Chloe left their campsite by canoe and paddle downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average speed of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream

Answers

Her average speed is 1.6 miles per hour. Average speed is total distance covered by total time taken to do it. She swims 4 miles upstream, and at 1 mph, it takes 4 hours. She comes back downstream at 4 mph and so she covers the 4 miles in 1 hour. Her total mileage is 8 miles. It takes 4 + 1 hours or 5 hours to cover it. The 8 miles divided by 5 hours is 1 3/5 miles per hour, or 1.6 mph for an average speed.

The time spent by the campers when they turn around downstream is 15 minutes.

Total distance traveled by Nick and Chloe

The concept of total distance traveled by Nick and Chloe can be used to determine the time they turn around downstream.

Let time for downstream = t1

Let time for upstream = t2

distance covered in upstream = distance covered in downstream = d

12(t1) = d

4(t2) = d

12t1 = 4t2

t1 + t2 = 1

t2 = 1 - t1

12t1 = 4(1 - t1)

12t1 = 4 - 4t1

16t1 = 4

t1 = 4/16

t1 = 0.25 hours

t1 = 0.25(60 min) = 15 mins

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A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m
and slope angle 30 degree, suppose the coefficient of friction of the slope surface is
0.25, find (neglect air resistance,take g=10 m/s^2)
The friction force experienced by the box.
00) The acceleration of the box along the slope.
(1) The time T required for the object to reach the bottom of the slope from the slope top.

Answers

Answer:

Explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7   And in order to answer what a is equal to, we find f:

f = μ[tex]F_n[/tex] where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:

[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and

[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.

A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?

Answers

Answer:

171.5J

Explanation:

K=1/2 *m *U²

K=1/2 *7 *7²

K=171.5 J

A 620 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 870 kg. As the elevator starts moving, the scale reads 450 N.

Required:
a. Find the acceleration of the elevator (magnitude and direction).
b. What is the acceleration if the scale reads 670 N?
c. If the scale reads zero, should the student worry? Explain.
d. What is the tension in the cable in parts (a) and (c)?

Answers

Answer:

(a) 9.28 m/s2

(b) 9.03 m/s2

(c) 9.8 m/s2

(d) 450 N, 670 N

Explanation:

mass of elevator + student, m = 870 kg

Reading of scale, R = 450 N

(a) When the elevator goes down, the weight decreases.

Let the acceleration is a.

By the Newton's second law

m g - R = m a

870 x 9.8 - 450 = 870 a

a = 9.28 m/s2

(b) R = 670 N

Let the acceleration is a.

870 x 9.8 - 670 = 870 a

a = 9.03 m/s2

(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.

(d) Tension in cable is 450 N and 670 N.

planet smaller than earth but larger than mercury​

Answers

Answer:

venus......................

If ATM is 102 kPa, what force does the atmosphere exert on the palm of your hand which has an area of 0.016 meters?

Answers

Answer:

Force = 1.632 Newton

Explanation:

Given the following data;

Pressure = 102 kPa

Area = 0.016 m²

To find what force the atmosphere exert on the palm of your hand;

Mathematically, pressure is given by the formula;

[tex] Pressure = \frac {Force}{area} [/tex]

Force = 102 * 0.016

Force = 1.632 Newton

What must be true if energy is to be transferred as heat between two bodies in physical contact?

1-The two bodies must have different volumes.

2-The two bodies must be at different temperatures.

3-The two bodies must have different masses.

4-The two bodies must be in thermal equilibrium.

Answers

Answer:

answer is d

Explanation:

i hope this helps you

A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?

Answers

Answer:

  T = 9056 K

Explanation:

In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation

                 λ T = 2,898 10⁻³

where lam is the wavelength of the maximum emission

                T = 2,898 10⁻³ /λ

let's calculate

                 T = 2,898 10⁻³ / 320 10⁻⁹

                  T = 9.056 10³ K

                  T = 9056 K

A boxer punches a sheet of paper in midair from rest to a speed of 20 m/s in 0.05 s. If the mass of the paper is 0.01 kg, the force of the punch on the paper is
A) 0.08 N.
B) 4.0 N.
C) 8.0 N.
D) 40 N.

Answers

The answer is: C) 8.0 N

a car runs of a road and collides with a tree. glass pieces from the windscreen are projected forward and are found an average distance of 12m from the car. the average height of the windscreen is 1.2m.


Establish the speed of the car at the time of impact. assume g=10 m/s²​

Answers

Answer:

v₀ₓ = 24.24 m / s

Explanation:

This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.

Y axis  

initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

when it reaches the ground its height is zero and the initial height is y₀=1.2m

          0 = y₀ + 0 - ½ g t²

          t = [tex]\sqrt{2 y_o/g}[/tex]

          t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]

          t = 0.495 s

         

X axis

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 12 / 0.495

          v₀ₓ = 24.24 m / s

1. A skydiver carries a buzzer which emits a steady 1800 Hz tone. A friend on the ground (directly below the skydiver) listens to the doppler shifted frequency. Assume the air is calm and that the speed of sound is independent of altitude. While the skydiver is falling at terminal velocity (the constant speed reached due to the balance of the downward gravitational force and the upward drag force due to air resistance), her friend on the ground hears waves of frequency 2150 Hz. a. How fast is the skydiver falling

Answers

Answer:

vs = 55.84 m/s

Explanation:

In this case, the source (skydiver with buzzer) is moving towards the observer (friend). The formula for this case will be:

[tex]f' = \frac{v}{v-v_s} f[/tex]

where,

f' = shifted frequency = 2150 Hz

f = actual frequency = 1800 Hz

v = speed of sound = 343 m/s

vs = speed of skydiver = ?

Therefore,

[tex]2150\ Hz = \frac{343\ m/s}{343\ m/s - v_s}(1800\ Hz)\\\\343\ m/s - v_s = \frac{343\ m/s}{2150\ Hz} (1800\ Hz)\\\\v_s = 343\ m/s - 287.16\ m/s\\[/tex]

vs = 55.84 m/s

A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.

Answers

Answer:

V = 240V

Explanation:

V = I*R

V = 15A*16ohms

V = 240V

A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?

Answers

Answer:

0.444atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (

P2 = final pressure (

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question,

P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm

P2 = ?

V1 = 80L

V2 = 160L (double of V1)

T1 = 34°C = 34 + 273 = 307K

T2 = 0°C = 0 + 273 = 273K

Using P1V1/T1 = P2V2/T2

0.999 × 80/307 = P2 × 160/273

79.92/307 = 160P2/273

Cross multiply

307 × 160P2 = 79.92 × 273

49120P2 = 21818.16

P2 = 21818.16 ÷ 49120

P2 = 0.444

P2 = 0.444atm

They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is filled with ice and water at equilibrium. What is the Carnot efficiency for their heat engine if the pressure is constant at 1.0 atmospheres?

Answers

Answer:

The efficiency of Carnot's heat engine is 26.8 %.

Explanation:

Temperature of hot reservoir, TH = 100 degree C = 373 K

temperature of cold reservoir, Tc = 0 degree C = 273 K

The efficiency of Carnot's heat engine is

[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]

The efficiency of Carnot's heat engine is 26.8 %.

What is the current in the 30 resistor?
A. 0.0833 A
B. 12 A
C. 2 A
D. 10 A

Answers

Answer:

Explanation:

Step 1) Combine all resistors into an equivalent overall resistor. These are all in series so you just add them up. Req = 10Ω + 20Ω + 30Ω = 60Ω:

Step 2) Using Ohm's Law, I = V/R = 120/60 = 2 A

Now you know how much current is flowing, and that current flows through each resistor the same. So the current in the 30 Ω resistor is 2.00 amps.

PLEASE HELP How does an object move when it is in linear motion?

in a straight line

up and down

in a circle

to the left

Answers

Answer:

In linear motion, the directions of all the vectors describing the system are equal and constant which means the objects move along the same axis and do not change direction. Correct Answer: In a straight line

Explanation:

Answer:

In a straight line. we can also have translational motion which is also a kind of linear motion .

What is this sport ⚽⚾

Answers

Answer:

sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.

hope it is helpful to you

Find the X and Y components of the following:
A. 35 m/s at 57q from the x-axis.

Answers

Explanation:

Given that,

35 m/s at 57° from the x-axis.

Speed, v = 35 m/s

Angle, θ = 57°

Horizontal component,

[tex]v_x=v\cos\theta\\\\=35\times \cos(57)\\\\=19.06 m/s[/tex]

Vertical component,

[tex]v_y=v\sin\theta\\\\v_y=35\times \sin(57)\\\\=29.35\ m/s[/tex]

Hence, this is the required solution.

Part C
When only one color of light reflects from a piece of paper, what happens to the other colors of light?
Remember that light is energy, and energy cannot be created or destroyed.
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Font Sizes
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Characters used: 0 / 15000
Plz help me yall I’m struggling

Answers

Answer:

the other colours get absorbed by the paper

Answer:

The other colors are absorbed by the paper and not reflected.

Consider a 92.0 kg ice skater who is spinning on the ice. What is the moment of inertia of the skater, if the skater is approximated to be a solid cylinder that has a 0.140 m radius and is rotating about the center axis of the cylinder.

Answers

Answer:

[tex]I=0.902kg*m^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=92.0kg[/tex]

Radius [tex]r=0.140m[/tex]

Generally the equation for moment of Inertia is mathematically given by

 [tex]I = 0.5*m*r^2[/tex]

 [tex]I=0.5*(92)*0.14^2[/tex]

 [tex]I=0.902kg*m^2[/tex]

Other Questions
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