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marin [14]
3 years ago
5

What value relates the speed of light in a vacuum to the speed of light in a given material? A) angle of incidence B) angle of r

efraction C) index of refraction D) index of reflection
Physics
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

C) Index of refraction

Explanation:

Index of refraction also known as refractive index is defined as speed of light in a vacuum divided by the speed of light in a given material.

The formula for Index of refraction is:

n = c ÷ v

where n represents Index of refraction,

c represents speed of light in a vacuum,

v represents speed of light in a given material.

The speed and direction of light changes when it moves from one medium to another. The change in direction is basically called refraction.

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
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Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

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a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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2 years ago
A glass rod and a steel rod are of equal length at 0C. At 100C they differ in length by
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The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

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l₀ = (l - l')/[(α- α')Δθ]

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l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

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Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

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