Answer:
Q = 4 Q₀
Explanation:
This is an exercise on capacitors, where the capacitance is
C =
if we apply the given conditions
C = \epsilon_{o} \ \frac{2A}{0.5d}
C = 4 \epsilon_{o} \ \frac{A}{d}
let's call the capacitance Co with the initial values
C₀ = \epsilon_{o} \ \frac{A}{d}
C = 4 C₀
The charge on each plate of a capacitor is
Q = C ΔV
If the potential difference is maintained, the new charge is
Q = 4 C₀ ΔV
let's call
Q₀ = C₀ ΔV
we substitute
Q = 4 Q₀
Answer:
(a) The distance of the chicken from the finish line is 62.5 m
(b) The stationary time of the wildebeest is 675 s
Explanation:
Given;
total distance traveled by wildebeest and chicken, d = 2 km = 2000 m
speed of the wildebeest, = 16 m/s
speed of the chicken, = 2.5 m/s
Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s
Time for chicken to finish the race without stopping, 2000/2.5 = 800 s
(b) for how long in time (in s) was the wildebeest stationary?
t(stationary) = t(chicken) - t(wildebeest)
t = 800s - 125 s
t = 675 s
(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?
The time taken for the wildebeest to run 1.6 km (1600 m) is given by;
t = 1600 / 16 = 100 s
The total time spent by the wildebeest before it resumed the race = stationary time + 100s
t (total) = 675 s + 100 s = 775 s
Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m
Thus, the distance of the chicken from the finish line = 2000 m - 1937.5 m
the distance of the chicken from the finish line = 62.5 m