A hockey puck with a mass of 160 g is sliding across an empty ice rink. The coefficient of friction between the hockey puck and the ice is 0.051. The puck travels 47.7 m before coasting to a stop. What was the initial velocity of the puck

Commer= truck drivertravel clerk=bachelors actuary= techinal program

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.

Anton because carbon dioxide and water are used with light energy to produce glucose sugar and oxygen gas.

Ur mom because that makes a lot of sense