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tatyana61 [14]
2 years ago
10

Which statement best describes the relationship between an object's density and its index of refraction? as the optical density

increases, the refractive index decreases. as the tendency of the atoms to absorb energy before reemitting it increases, the refractive index increases. as the mass per unit volume of the substance increases, the refractive index decreases. as the physical density of the substance increases, the refractive index increases.
Physics
2 answers:
tino4ka555 [31]2 years ago
7 0

Answer:

as the tendency of the atoms to absorb energy before reemitting it increases, the refractive index increases.

Explanation:

The process of being absorbed and re-emitted by the atoms of the material decreases the speed of light from one end of the medium to the other. Therefore, the higher the absorption, the lower its speed in any medium with respect to the speed of light in a vacuum. The index of refraction indicates how slow the light in the material is compared to its speed in a vacuum. Thus, if the atoms absorb more energy before reemitting it increases, the refractive index increases

babunello [35]2 years ago
4 0
For future references, the answer is B.
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A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
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Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, 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

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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 \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}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) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

8 0
3 years ago
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