Answer:
Minimum capacitance = 200 μF
Explanation:
From image B attached, we can calculate the current flowing through the capacitors.
Thus;
Since V=IR; I = V/R = 5/500 = 0.01 A
Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V
So minimum capacitance will be determined from;
i(t) = C(dv/dt)
So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]
C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF
-- If the frequency of a wave is too high for our eyes to detect it ... but not <u>too too</u> high ... we call it an <em>ultraviolet </em>wave.
-- If the wave's frequency is even higher than that, we call it an <em>X-ray</em> wave.
-- If the wave's frequency is even higher than that, we call it a <em>Gamma Ray </em>wave.
The watts determine the brightness. Watt is the unit of Power. And Power is equal to Voltage x Amps (current). So both the current and voltage determine the brightness.
Answer:
0.06 A
Explanation:
We have given mass =0.954 kg
velocity =1.27 m/sec
efficiency =0.361 the output kinetic energy of the motor 
Efficiency =0.361
so input to the motor = output/efficiency
so input to the motor = 
we know that
where E is energy and T is time so 
We know that power P=VI we have given V=4.5 VOLT
So current 