Suppose you are given a 4 m long straight bar of uniform density balanced at its center and a weight of 4.6 kg hanging on the bar at a distance of 12.0 cm from the pivot. Where must we hang a 2.6 kg mass to have the system remain in balance?

21.23 cm

Explanation:

Being that the mass of the bar is equal on both sides, the moment produced by the masses cancel each other out, so you can just look at the hanging masses.

Sum of the moments about the fulcrum = 0... I'll say that moments produced on the 4.6 kg side are positive

0 = 4.6kg * g * 12.0 cm - 2.6kg * g * x

g's cancel. solve for x

x =( 4.6  * 12) / 2.6

x = 55.2 / 2.6

x = 21.23cm

so 2.6kg mass should be pivoted at a distance of 21.23 cm from pivot.

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.