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Mkey [24]
3 years ago
15

A ray moving in plastic at 62.9 deg enters water, where it bends to 70.9 deg. What is the index of refraction of the plastic?

Physics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

refractive index of plastic is 1.42

Explanation:

When light ray enters from one medium to other medium then due to transition of light it bends away or towards the normal, this phenomenon is known as refraction of light

So here we know that

n_1 sin i = n_2 sin r

here we have

n_2 = \frac{4}{3}

i = 62.9^o

r = 70.9^o

now we have

n_1 sin62.9 = (\frac{4}{3}) sin70.9

n_1 = 1.42

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Answer:

d. We can calculate it by applying Newton's version of Kepler's third law

Explanation:

The measurements of a Star like the Sun have several problems, the first one is distance, but the most important is the temperature since as we get closer all the instruments will melt. This is why all measurements must be indirect because of the effects that these variables create on nearby bodies.

Kepler's laws are deduced from Newton's law of universal gravitation, in these laws the mass of the Sun affects the orbit of the planets since it creates a force of attraction, if measured the orbit and the time it takes to travel it we can know the centripetal acceleration and with it knows the force, from where we clear the mass of the son.

Let's review the statements of the exercise

.a) False. We don't have good enough models for this calculation

.b) False. The size of the sun is very difficult to measure because it is a mass of gas, in addition the density changes strongly with depth

.c) False. The amount of light that comes out of the sun is not all the light produced and is due to quantum effects where the mass of the sun is not taken into account

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Correct answer is D

6 0
3 years ago
Can someone help me?
Alik [6]

Answer:

Resultant = 3.05N

Explanation:

r =  \sqrt{{5}^{2} + 5^{2} - 2(5)(5) \cos(120) }

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6 0
2 years ago
If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?
Anon25 [30]
<h2>Answer:50ms^{-1}</h2>

Explanation:

Let v_{a} be the airspeed.

Let v_{w} be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If v_{1} and v_{2} are two perpendicular vectors,the resultant vector has the magnitude \sqrt{|v|_{1}^{2}+|v|_{2}^{2}}

Given,

v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}

So,the ground speed is \sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}

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