An 23,000kg asteroid passes by within 5m of a 325,000kg spaceship. What is the force due to gravity between the asteroid and the spaceship?
0 149 N
0 627 N
0 0.02N
0 0.08N

Bsame force but in the same direction

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.

Fb is the magnetic force  q is the chargev is the velocity of the charge  b is the magnetic field  the equation is: f=|q|vbsin0