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dlinn [17]
3 years ago
10

The most common type of mirage is an illusion that light from faraway objects seem to be reflected by a pool of water that is no

t really there. Mirages are generally observed in deserts, when there is a hot layer of air near the ground. The index of refraction of air is lower at higher temperature. What physical phenomena explains how mirages are formed? Give a brief explanation of the phenomena as your LPE.
Physics
1 answer:
olga nikolaevna [1]3 years ago
5 0

Answer:

Mirage is an optical illusion due to the refraction of light as it travels through different layers of air at different temperature.

Explanation:

On a clear hot afternoon, the sun heats up the ground, and the ground heats up the air immediately above it. The result is that the air directly above the ground is hotter than the air above it, leaving the air layers with different refractive indies. Since refractive index decreases with increase in temperature, the air immediately above the ground bends the light that travels through it from the sky, directly into our eyes. The image seen is perceived to be reflected from the ground, but in an actual sense is a direct image of the sky, giving one the impression of seeing a pool of water on the ground.

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Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

          m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

where,  m_{1} = mass of the projectile

            m_{2} = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

         75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v

                                  = \frac{45}{50.075}

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E = \frac{1}{2}mv^{2}

                  = \frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}

                  = 13500 J

Now, energy after the impact will be as follows.

             E' = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

           \Delta E = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n = \frac{\Delta E}{E}

             = \frac{13480}{13500} \times 100

             = 99.85

             = 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

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