calculate the force of gravitational attraction between the earth (5.98 x 10^24kg) and a 65-kg passenger who's in an airplane at 40000 feet above the earth's surface. this would place the passenger a distance of 6.39 x 10^6m from the earth's center.

1point) a skydiver weighing 130 lb (including equipment) falls vertically downward from an altitude of 14,000 ft and opens the parachute after 20 s of free fall. assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude 0.85|v|0.85|v| when the parachute is closed and is of magnitude 17|v|17|v| when the parachute is open, where the velocity vv is measured in ft/s. assume that acceleration due to gravity has magnitude 32 ft/s/s; remember that weight is the product of mass and gravitational acceleration. find the speed of the skydiver when the parachute opens. speed is 150.6121788 ft/s find the skydiver's height when the parachute opens. height is 2338.98591 ft find the limiting velocity vlvl after the parachute opens. (think about whether this should be positive or negative.) vl=vl= 7.647 ft/s

What properties of a moving object are used in determining the object's energy of motion

What causes the formation of ocean currents?