The period of a simple pendulum that is 1.00m long on another planet is 1.7s. What is the acceleration due to gravity on this planet if the mass of the pendulum bob is 1.5kg?

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.

There was no intial y-momentum so the net final y-momentum must also be 0. they can have final velocities in the y-direction

Two 10¢  coins  (dimes)  carrying identical charges  are  lying 2.5 m apart  on a  table.  if each  of  these coins experiences  an  electrostatic force  of  magnitude 2.0 n due  to the  other coin.  the charge on each coin is :     52.2 uc