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yawa3891 [41]
2 years ago
13

The meaninn of dynamics​

Physics
1 answer:
Studentka2010 [4]2 years ago
6 0

Answer:

Dynamics is just a nice word used in physics denoting a branch of physics, related to the study of forces. Usually these forces are not in mechanical equilibrium, else the branch would be statics.

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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
Where is the Oort cloud located?
sergiy2304 [10]

Answer: C

Explanation: just did it

8 0
3 years ago
Read 2 more answers
Momentum and different surfaces
saw5 [17]
What are you asking here?
5 0
3 years ago
1 point
Novay_Z [31]

Answer:

Explanation:

Not sure what your options are but anything that says something like

"at the block surface in contact with the ramp along the line from V to Z"  is probably a good shot.

4 0
2 years ago
Jamal is at Six Flags playing at the arcade. At one booth he throws a 0.50 kg ball
Mekhanik [1.2K]

were solving for v velocity of the ball after it has hit the bottle. a. momentum ->p=mv->ball + bottle momentum during hit = ball + bottle momentum after hit-> ball (.5*21)+ bottle (.2*0) (it's 0 because the the bottle is standing still) = ball after hit (.5*v)+bottle after hit (.2*30) -> 10.5+0=.5v+6 ->4.5=.5v->v=9m/s

b. if bottle was heavier the ball would be slower so final velocity would decrease

8 0
2 years ago
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