Help me solve this physics homework:
in sense we all have kinetic energy,even when we are standing still.the earth with a radius of 6.37*10power6 m,rotates about its axis once a day.ignoring the earths rotation about the sun,what is the K.E of a 50Kg man standing on the surface of the earth.
It will depend upon the longitude where he is standing. The velocity of a person at the equator, in an Earth-fixed coordinate system, is
V = 2*pi*6.37*10^6 m/(24*3600 sec)
= 463 m/s
Use that velocity with the formula for kinetic energy, (1/2)MV^2