A Hooke's law spring is compressed a distance d and is used to launch a particle of mass m vertically to a height h above its starting position. Under double the compression, the spring is now used to launch a particle of mass 2 m. How high does the second mass rise above its starting position

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Answerapplying a direct current to a non magnetic material in order to create a electromagnetic field; plz check first good luck;

False  i read once that you can't do more than two activities at once