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Anna007 [38]
3 years ago
11

Explain how you can cause light separated by a prism to combine

Physics
1 answer:
r-ruslan [8.4K]3 years ago
7 0
When you shine a lite through a prism is reflects out light through all of the edges and causes light separation. Or just simply shine a laser through the edge of a sideways piece of glass.

I hope that this was helpful for you.
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What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.
Pavel [41]
Hopefully this will help you.

4 0
3 years ago
Please help!
den301095 [7]

Hello there,

It takes 300 newtons of force and a distance of 20 meters for a moving cart to come to a stop. How much kinetic energy did this cart have?

Answer:  6000


8 0
3 years ago
g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the
ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

        a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength  λ = 630 nm = 630 10⁻⁹ m and the width of the slit  a = 5 10⁻⁶ m

         sin  θ = m λ / a

         sin  θ = 3 630 10⁻⁹ / 5 10⁻⁶

         sin  θ = 3.78 10⁻¹ = 0.378

          θ = sin⁻¹  0.378

         

to better see the result let's find the angle in radians

          θ = 0.3876 rad

let's reduce to degrees

         θ = 0.3876 rad (180º /π rad)

         θ = 22.2º

4 0
3 years ago
A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t
const2013 [10]

Answer:

\theta=12.19^{\circ}

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

sin\theta =\dfrac{u}{v}

Now by putting the values in the above equation we get

sin\theta =\dfrac{30}{142}

sin\theta=0.21

\theta=12.19^{\circ}

Therefore the angle will be 12.19° .

 

4 0
3 years ago
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