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gladu [14]
3 years ago
6

A light wave has a 670 {\rm nm} wavelength in air. Its wavelength in a transparent solid is420 {\rm nm} .a)What is the speed of

light in this solid?b)What is the light's frequency in the solid?
Physics
1 answer:
tia_tia [17]3 years ago
5 0

Answer:

Explanation:

given,

Light wave wavelength in Air, λ_(vac) = 670 nm

wavelength in transparent solid, λ= 420 nm

Refractive index of the solid

n = \dfrac{\lambda_{vac}}{\lambda}

n = \dfrac{670}{420}

   n = 1.6

a) speed of light in the solid

v = \dfrac{c}{n}

v = \dfrac{c}{1.6}

       v = 0.625 c

b) light's frequency in the solid

  f = \dfrac{c}{\lambda_{vac}}

frequency of light remain same when light move from one medium to another.

 f is the frequency of light

  f = \dfrac{3\times 10^8}{670\times 10^{-9}}

        f = 4.47 x 10⁻¹⁴ Hz

the light's frequency in the solid is equal to 4.47 x 10⁻¹⁴ Hz

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mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

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Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

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Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

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            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

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Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

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