9. An ore cart of mass 1200 kg is rolling at a speed of 10.8 m/s across a flat surface. A crane dumps 496 kg of ore into the cart from directly above it. How fast does the cart move after being loaded with ore? Assume that frictional forces with the flat surface are negligible.
O 6.9 m/s
O 7.6 m/s
O 4.6 m/s
5.1 m/s

Abecause its more effective

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.

C. the electric current passes from the slip rings, through the brushes, and to the collection device.