The earth's radius is 6.37×10^6m; it rotates once every 24 hours. with the angular speed of 7.3 x 10^-5 what is the speed of a point on the earth's surface located at 2/3 of the length of the arc between the equator and the pole, measured from equator? (hint: what is the radius of the circle in which the point moves? )

The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.

Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.

r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6

v - 232.5 m/s

1) when you rub your hands together quickly, the kinetic energy is converted to mechanical energy.2) thermal energy from the sun travels to earth as infrared waves.3) energy is transferred through the matter of space by electromagnetic waves. 4) when a match is burned, chemical potential energy is converted to thermal and radiant energy. 5) seismicwaves are caused by back and forth movement , or vibration, of an object. 6) mechanicalenergy in an engine changes thermal energy into kinetic energy to make a car move.