A 50.0 kg child is riding in a 120.0 kg bumper car at an amusement park. The child drives the bumper car at 4.5 m/s head-on into the wall of the ride and rebounds at 1/3 the speed. What is the magnitude of the change in momentum of the child in the bumper car? A. 375 kg•m/s
B. 510 kg•m/s
C. 720 kg•m/s
D. 1020 kg•m/s

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

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf,  ℰ,  to a battery's terminal voltage, v, current, i, and internal resistance, r, is    ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

Ibelieve the answer is d. macula  hope my answer has you : )

It is the areas with lines. this may sound a little confusing, but i hope it !