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goldfiish [28.3K]
3 years ago
11

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible

light ranges in wavelength from 400 nm (violet) to 700 nm (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of:
(a) The wavelength inside the vitreous humor?
(b) The frequency?
Physics
1 answer:
Leto [7]3 years ago
4 0

Answer:

a)   298.5 nm ,  522.4 nm  and b)  radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

    λₙ = λ₀ / n

    λₙ = 400 nm    in the void

    λₙ = 400 / 1.34

    λₙ= 298.5 nm

   λ₀ = 700 nm

   λₙ = 700 / 1.34

   λₙ = 522.4 nm

The radiation frequency does not change

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The motion of the ceiling is y = Y sinωt

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= \sqrt { \frac{100*32.2}{200}}

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X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

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4 0
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9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

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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

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

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

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the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

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          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

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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

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
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