Unpolarized light of intensity i0 is incident on three polarizing filters. the axis of the first is vertical, that of the second
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:
the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Explanation:
The computation of the magnitude of the each force is shown below:
Provided that
Ratio of forces = 3: 5
Let us assume the common factor is x
Now
first force = 3x
And, the second force = 5x
Resultant force = 35 N
The Angle between the forces = 60 degrees
Based on the above information
Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos
35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]
35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)
35 = √49 x²
x = 5
So, the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
1) Distance down the hill: 1752 ft (534 m)
2) Time of flight of the shell: 12.9 s
3) Final speed: 326.8 ft/s (99.6 m/s)
Explanation:
1)
The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:
(1)
where:
is the initial vertical velocity of the shell, with
and
is the acceleration of gravity
At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation
where is the initial horizontal velocity of the shell.
We can re-write this last equation as
(1b)
And substituting into (1),
(2)
where we have choosen the top of the hill (starting position of the shell) as origin (0,0).
We also know that the hill goes down with a slope of from the horizontal, so we can write the position (x,y) of the hill as
(3)
Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):
Substituting (1b) into this equation,
Which has 2 solutions:
x = 0 (origin)
and
So, the distance d down the hill at which the shell strikes the hill is
2)
In order to find how long the mortar shell remain in the air, we can use the equation:
where:
x = 1322 ft is the final position of the shell when it strikes the hill
is the initial velocity of the shell
is the angle of projection of the shell
Substituting these values into the equation, we find the time of flight of the shell:
3)
In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.
The horizontal component of the velocity is constant and it is
Instead, the vertical component of the velocity is given by
And substituting at t = 12.9 s (time at which the shell strikes the hill),
Therefore, the final speed of the shell is:
Learn more about projectile motion:
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