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sesenic [268]
4 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]4 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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Answer: Your question does not make sense

Explanation:

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4 years ago
What form of energy is released into the atmosphere by the earth's surface
Crank

Answer:

Thermal Energy (Heat)

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3 years ago
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-<br> ZOOLS<br> 6) The mass of a motorcycle is 250 kg. What is?<br> A) Its weight on Earth in Newtons?<br> B) Its weight on the
marishachu [46]

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

4 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
When an object accelerates, what about its motion changes? Question 1 options: Speed must change, but not velocity. Both speed a
QveST [7]

Answer:

The velocity must change but not speed.

Explanation:

  • Velocity is defined as the displacement by time. Whereas speed is expressed as the distance between two successive positions of the body to the time interval it took to travel.

                <em>Velocity,        V = D / t        m/s</em>

<em>                  Speed,          s = d /t          m/s        </em>  

  • Velocity is a vector quantity that has a magnitude and direction.
  • The speed is a scalar quantity having only the magnitude.
  • At any instant of time, the magnitude of the velocity is always equal to the magnitude of the speed. The magnitude of velocity, |<em>v </em>| = magnitude of speed, |<em>v </em>|. The magnitude is always positive
  • The acceleration of a body is defined as the rate of change of velocity to time.

                               <em>   a = (v - u) / t      m/s²</em>

  • If a body is accelerating, It varies its velocity with respect to time.
  • In case of uniform circular motion, the speed remains constant, but the velocity changes continuously.

So, in the case of circular motion if an object accelerates, velocity must change but speed remains constant.

5 0
4 years ago
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