The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.
Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.
Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
Answer:
reliability
accuracy
Explanation:
If a reading of a measurement is consistently the same then the measurement is reliable.
If a reading of measurement is close the actual value of the measurement then the reading is accurate.
Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.
Hence, the radar has problems in its reliability and accuracy
Answer:
75 rotations
Explanation:
f0 = 0, f = 3000 rpm = 50 rps, t = 3 s
(a) use first equation of motion for rotational motion
w = w0 + α t
2 x 3.14 x 50 = 0 + α x 3
α = 104.67 rad/s^2
(b) Let θ be the angular displacement
use second equation of motion for rotational motion
θ = w0 t + 1/2 α t^2
θ = 0 + 0.5 x 104.67 x 3 x 3
θ = 471.015 rad
The angle turn in one rotation is 2 π radian.
Number of rotation = 471.015 / (2 x 3.14) = 75 rotations
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.