An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 28.9 m/s . It then flies a further distance of 45300 m , and afterwards, its velocity is 47.5 m/s . Find the airplane's acceleration.

a = 0.016 m/s^2

the airplane's acceleration is 0.016 m/s^2

Explanation:

Applying the equation of motion;

v^2 = u^2 + 2as 1

Where;

v = final velocity = 47.5 m/s

u = initial velocity = 28.9 m/s

a = acceleration

s = displacement = 45300m

From equation 1, making a the subject of formula;

v^2 = u^2 + 2as

2as = v^2 - u^2

a = (v^2 - u^2)/2s

Substituting the given values;

a = (47.5^2 - 28.9^2)/(2×45300)

a = 0.015684768211 m/s^2

a = 0.016 m/s^2

the airplane's acceleration is 0.016 m/s^2

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(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.