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WARRIOR [948]
3 years ago
12

Light enters water from air at an angle of 25° with the normal, Θ1. If water has an index of refraction of 1.33, determine Θ2.

Physics
1 answer:
vladimir1956 [14]3 years ago
6 0
By Snell's law:

η = sini / sinr.        i = 25,  η = 1.33

1.33 = sin25° / sinr

sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177    Use a calculator.

r = sin⁻¹(0.3177)

r ≈ 18.52°

Option A.

God's grace.
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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
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Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

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