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Marta_Voda [28]
3 years ago
14

Which index of refraction represents the most optically dense material?

Physics
2 answers:
faust18 [17]3 years ago
6 0
The relation between the refractive index and the optical density of the material is a direct relation.
This means that the more the refractive index is, the more optically dense the material is.

Based on the above, when checking the given choices, the refractive index that represents the most optically dense material would be the largest refractive index which is:
<span>d. 2.65</span>
Tju [1.3M]3 years ago
6 0

Answer:

Explanation:It does ask for the most optically dense NOT the least.

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Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

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comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

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  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

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