Acart's initial velocity is +3.0 meters per second. what is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0 seconds?

You would need to take a picture so i can see

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

Following the american gauge wire standards ,the larger gauge number, the slimmer the wire and thus the smaller current carrying capacity it can have.then both statements would be incorrect, as they are saying the opposite of what it is actually is.(smaller gauge numbers have thicker wires and can carry more current)