Carbon and silicon are found in the earth's crust and in what type of bonds?
1 answer:
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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.
Answer:
v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²
Explanation:
The expression left corresponds to an oscillatory movement (MAS), the speed is defined by
v = dr / dt
the function of position
r = 2 cos πt i^ + 3 sin πt j^
let us note that it is a movement in two dimensions
let's perform the derivative
v = -2π sin πt i^ + 3π cos πt j^
we evaluate this expression for t = 0.25 s, remember that the angle is in radians
v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^
v = (-4.44 i^ + 6.66 j^ ) m/s
To calculate the mean acceleration we use the expression
a = () / Δt
indicates that the time is the first 3 s
we look for the initial velocity t = 0 s
v₀ = 0 i ^ + 3π j ^
we look for the fine velocity, t = 3 s
v_f = - 2π sin (π 3) + 3π cos (π 3) j ^
v_f = 0 i ^ - 3π j ^
we calculate the average acceleration
Δt = (3 -0) = 3 s
a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3
a_average = (0 i ^ -2π j ^ ) m/s²