Answer:
2.9*10^14 electrons
Explanation:
An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.
An electron has a charge of 1.6*10^-19 C.
35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons
So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.
Answer:
Hello your question lacks the required diagram attached below is the missing diagram
final compensator = k ( s + 9.3731 )
gain = 695.4275
Explanation:
attached below is the remaining part of the detailed solution of the question
C(s) =
The dominant poles fn 20.5% are at Sd = -2.91 + 5.77 J
uncompensated system setting time (ts) = 4/2.91 = 1.3746