9. A piece of an unknown metal with a mass of 21.7 g is heated to 100.0°C and dropped into 50.0 mL of water at 23.6°C. The final temperature of the system is 32.5°C. What is the specific heat capacity of
the metal?

The magnets would be | n s | and | s n|

Sorry i don’t know what to do

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

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.