A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.9 m/s (about 78 mph) and the ball is 0.310 m from the elbow joint, what is the angular velocity of the forearm

112.581 rad/s.

Explanation:

Angular velocity: This can be defined as the velocity of a body undergoing angular motion. The S.I unit of angular velocity is rad/s.

From the question, The expression for angular velocity is given as,

ω = v/r Equation 1

Where ω = angular velocity of the forearm, v = velocity of the ball in the pitcher's hand, r = the distance of the ball from the elbow joint.

Given: v = 34.9 m/s, r = 0.310 m

Substitute into equation 1

ω = 34.9/0.310

ω = 112.581 rad/s.

Hence the angular velocity of the forearm = 112.581 rad/s.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

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.