The battery will be full still a 8v bc of no time comparison
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
Capacitance, C = 26.1 picofarad
Explanation:
It is given that,
Side of square, x = 4.3546 cm = 0.043546 m
Distance between electrodes, d = 0.6408 mm = 0.0006408 m
Voltage, V = 73.68 V
Capacitance of parallel plates is given by :
or
C = 26.1 picofarad
So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.
The car will speed up because if the air resistance and friction forces are reacting in the direction opposite to the Applied force, The Net force will still be in the direction of the car's motion and since F=ma a will cause your velocity to increase
