If you push a crate horizontally with 100 N across a 10-m factory floor and the friction between the crate and the floor is a steady 70 N, how much kinetic energy does the crate gain?

300J

Explanation:

Parameters given:

Force applied on crate, F = 100N

Distance moved by crate, D = 10m

Frictional force, Fr = 70N

The work done by pushing the crate is given by:

W = F * D

= 100 * 10 = 1000J

The work done by friction is given as:

W(Fr) = -Fr * D

= -70 * 10 = -700J

Work done by friction is negative because it works against the motion of the crate.

Net work done = 1000 + (-700)

WD = 1000 - 700 = 300J

Work done is equivalent to the change in kinetic energy.

WD = ∆KE

∆KE = KE(final) - KE(initial)

KE(initial) = 0

Therefore,

WD = KE(final)

=> KE(final) = 300J

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(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.