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strojnjashka [21]
3 years ago
15

A tennis ball is dropped from 1.62 m above the

Physics
1 answer:
natita [175]3 years ago
8 0

Answer:

-5.63 m/s

Explanation:

Given:

y₀ = 1.62 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.62 m)

v = -5.63 m/s

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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

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(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

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Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

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(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

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