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tatyana61 [14]
3 years ago
5

Consider light energy that is momentarily absorbed in glass and then re-emitted. Compared to the absorbed light, the frequency o

f the re-emitted light is
A) considerably less.
B) slightly less.
C) the same.
D) slightly more.
E) considerably more.
Physics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

Compared to the absorbed light, the frequency of the re-emitted light is <em>the same</em>.

Explanation:

The reason to why the light energy that is momentarily absorbed in glass and then re-emitted, compared to the absorbed light, the frequency of the re-emitted light is <em>the same</em> is because the wavelength of the absorbed light and the re-emitted light is the same. The energy carried by an electromagnetic wave is proportional to the frequency of the wave. The wavelength and frequency of the wave are connected via the speed of light:

<em>c = fλ</em>

<em>where</em>

  • <em>c is the speed of light</em>
  • <em>f is the frequency of light </em>
  • <em>λ is the wavelength of the light</em>

Though light does not have mass, it does have energy and it's energy is conserved. Therefore, in the conservation of energy, the energy of the absorbed light must be equal to the re-emitted light. This is why the frequency of the re-emitted light is <em>the same</em> as the frequency of the absorbed light.

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Explanation:

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6 0
2 years ago
block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

7 0
3 years ago
Which are characteristics of electromagnetic waves?check all that apply.
emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces  --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light  --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, c=3\cdot 10^8 m/s)

travel only through matter  --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space  --> FALSE, electromagnetic waves can also travel through matter

can bend around objects  --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other  --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave

8 0
3 years ago
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Longitudional waves travel through a series of ________ and ___________.
nekit [7.7K]

Answer:

compressions; rarefactions

Explanation:

4 0
3 years ago
A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy
Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis  

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be F_1 and force on +Q charge y axis due to -Q charge on x-axis be F_2.

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = \sqrt{2}d units

k_e= Coulomb constant

F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N

F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N

Net force on +Q charge on y-axis is:

F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N

F_y=F_1-F_2cos45^o

F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})

F_N=\sqrt{F_x^2+F_y^2}

|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|

The net froce on the +Q charge on y-axis is

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

4 0
3 years ago
Read 2 more answers
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