Ajet airliner moving initially at 548 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 343 mph in a direction 67◦ north of east. what is the new speed of the aircraft with respect to the ground? answer in units of mph.

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

along the x-direction

in a direction north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

So, the components of the resultant velocity in the two directions are

So the new speed of the aircraft is:

i need to show me the graph so i can give u an answer

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.