Derive the expression ε=ln(1+e), where ε is the true strain and e is the engineering strain. Note that this expression is not va
1 answer:
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Answer:
The answer is below
Explanation:
a) The work done is equal to the loss in kinetic energy (KE)
Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy
Final KE =
But the final velocity is 0 (at rest). Hence:
Final KE =
ΔKE = 0 - K = -K
W = ΔKE = -K
Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)
W = qEd
but q = -e, hence:
W = -e * E * d
Using:
W = ΔKE
-e * E * d = -K
E= K / (e * d)
b) The electric field is in the direction of the electrons motion
Answer:
2.35 + j8.34 Ω
Explanation:
Voltage = V = 240 V rms
supplying power = S = 8 kVA
power factor = pf = 0.6
Let P₁ represents one load draws 3kW at unity powder factor
The power angle is:
θ = cos⁻¹ pf
= cos⁻¹ 0.6 = 53.13°
Complex power supplied source is:
S = S
< θ
= 8<53.13° kVA
Complex power for first load:
S₁ = P₁ = 3kVA
Since the power angle of first load is θ₁ = 0°
According to principle of conservation of AC power, the power of second load is:
S₂ = S - S₁
= 8<53.13° - 3
= 6.65<74.29° kVA
Since the second load is a Y connected load the phase voltage:
V = V
/
= 240/1.732051
= 138.564
= 138.56 V
Complex power of second load:
S₂ = 3 V² / Z
impedance per phase of the second load:
Z = 3 V
² / S₂
= 3 (138.56)² / 6.65<74.29°
= 3(19198.8736) / 6.65<74.29°
= 57596.6208 / 6.65<74.29°
Z = 2.35 + j8.34Ω