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fomenos
2 years ago
5

Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a wa

ter layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:
Physics
1 answer:
creativ13 [48]2 years ago
3 0

Answer:

in oil film        λ = 303.57 10⁻⁹ m

in the water film    λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

             v = λ f

in the void we have

             c = λ₀ f

we divide the two expression

            c / v = λ₀ / λ

the refractive index is

             

              n = c / v

              n = λ₀ /λ

              λ = λ₀ / n

let's calculate

in oil film

            λ = 425 10⁻⁹ / 1.40

            λ = 303.57 10⁻⁹ m

in the water film

            λ = 425 10⁻⁹ / 1.33

            λ = 319.55 10⁻⁹

those wavelengths are in the ultraviolet

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A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

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<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

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To get the position equation we just need to integrate the above equation:

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$d u=-\sin (\omega t) d t

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Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

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$f(t)=\frac{\cos ^3(\omega t)}{3}+C

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1 year ago
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8 0
3 years ago
The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr
Dvinal [7]

Answer:

T_A_v_g=9.918192559$^{\circ}C

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

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The average temperature over a given interval can be calculated as:

T_a_v_g=\frac{T_o+T_f}{2}

Where:

T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

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Hence, the average temperature between noon and midnight is:

T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C

3 0
3 years ago
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