A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
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Answer:
The tangential speed of the tack is 8.19 m/s.
Explanation:
The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:
Use the relation of angular speed with tangential speed to find the tangential speed of the tack.
The tangential speed v of the tack is given by following expression
v = ω r
Here, r is the distance to the tack from axis of rotation.
Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.
v = 21.174 × 0.387
v = 8.19m/s
Thus, The tangential speed of the tack is 8.19 m/s.
The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:
We can clear time in the speed equation:
If you find some mistake in my English, please tell me know.
We can clear time in the speed equation:
If you find some mistake in my English, please tell me know.
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:
(1)
Where 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):
(2)
Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:
(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:
(4)
Case for the object with the blackbody emission spectrum peak in the blue:
Before replacing all the values in equation (4), (450 nm) will be express in meters:
⇒
Case for the object with the blackbody emission spectrum peak in the red:
Following the same approach above:
⇒
Comparison:
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.
Check the attached file for the answer.
