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sesenic [268]
3 years ago
13

nsider a common mirage formed by superheated air immediately above a roadway. A truck driver whose eyes are 2.00 m above the roa

d, where n = 1.000 293, looks forward. She perceives the illusion of a patch of water ahead on the road. The road appears wet only beyond a point on the road at which her line of sight makes an angle of 1.30° below the horizontal. Find the index of refraction of the air immediately above the road surface. (Treat this as a problem in total internal reflection. Give your an
Physics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:

n_{air}  =1.0000355

Explanation:

from snells law we have following relation

n' Sin i = n_{air} Sin r


n_{air} is refractive index of air

n' -  refractive index of air below

i is angle of incidence = 90 - 1.30 = 88.70^o

r is refraction angle

plugging all value in above formula

1.000293 \times sin 88.7 = n_{air} \times sin 90

n_{air}  =1.0000355

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Describe the composition of humus and why it is an effective organic fertilizer
kenny6666 [7]

Answer:

soil additive

Explanation:

Soil Additive

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2 years ago
How far away is mars?
Elanso [62]

Answer:

about: 110.14 million mi

Explanation:

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Hope that was helpful.Thank you!!!

3 0
3 years ago
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a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
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A. is the right answer since work is negative and Q which is heat in negative also

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Acceleration is the force of something moving, or acceleration can be how fast an object is going!
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