What is the connection between electrons and protons
1 answer:
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Answer:
Explanation:
Let the angle between the first polariser and the second polariser axis is θ.
By using of law of Malus
(a)
Let the intensity of light coming out from the first polariser is I'
.... (1)
Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.
By the law of Malus
So,
(b)
Now differentiate with respect to θ.
Answer:
Option d
The minimum angular separation between two objects that the Hubble Space Telescope can resolve is .
Explanation:
The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).
That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).
The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.
The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.
The angular resolution can be determined in analytical way by means of the Rayleigh criterion.
(1)
Where is the wavelength and D is the diameter of the telescope.
Notice that it is necessary to express the wavelength in the same units than the diameter.
⇒
Finally, equation 1 can be used.
Hence, the minimum angular separation between two objects that the Hubble Space Telescope can resolve is .
Answer:
The answer to the question
The steady state response is u₂(t) = -cos(3t + π/4)
of the form R·cos(ωt−δ) with R = , ω = 3 and δ = -π/4
Explanation:
To solve the question we note that the equation of motion is given by
m·u'' + γ·u' + k·u = F(t) where
m = mass = 2.00 kg
γ = Damping coefficient = 1
k = Spring constant = 3 N·m
F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)
Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)
The homogeneous equation 2·u'' + u' + 3·u is first solved as follows
2·u'' + u' + 3·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by =
This gives the general solution of the homogeneous equation as
u₁(t) =
For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)
Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have
3·B-15·A = 27
3·A +15·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)
u₂(t) = 1.5·(sin(3·t) - cos(3·t) = ·cos(3·t + π/4)
The general solution is then u(t) = u₁(t) + u₂(t)
however since u₁(t) = ⇒ 0 as t → ∞ the steady state response = u₂(t) =
·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where
R =
ω = 3 and
δ = -π/4