a) precise but not accurate.
this is because the thrower was capable to repeat the same result everytime but it was not the desire outcome.
Answer:
The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.
Answer:
The frictional force acting on the bear during the slide is 207.5 N
Explanation:
Given;
mass of beam, m = 25-kg
vertical height, h = 12 m
speed of fall, v = 6 m/s
Change in potential energy of the beam:
ΔP.E = -mgh = - 25 x 9.8 x 12 = -2940 J
Change in kinetic energy of the beam:
Δ K.E = ¹/₂mv² = ¹/₂ x 25 x (6)² = 450 J
Change in thermal energy of the system due to friction:
ΔE = - (ΔP.E + Δ K.E)
ΔE = - (-2940 J + 450 J)
ΔE = 2940 J - 450 J = 2490 J
Frictional force (in N) acting on the bear during the slide:
F x d = Fk x h = ΔE
Where;
Fk is the frictional force
Fk = ΔE/h
Fk = 2490J / 12m
Fk = 207.5 N
Therefore, the frictional force acting on the bear during the slide is 207.5 N
