Answer:
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Answer:
![\triangle V_0=0.08V](https://tex.z-dn.net/?f=%5Ctriangle%20V_0%3D0.08V)
Explanation:
From the question we are told that:
Incremental resistance ![R=8ohms](https://tex.z-dn.net/?f=R%3D8ohms)
Resistor Feed ![R_f=82ohms](https://tex.z-dn.net/?f=R_f%3D82ohms)
Supply Change ![\triangle V=1](https://tex.z-dn.net/?f=%5Ctriangle%20V%3D1)
Generally the equation for voltage rate of change is mathematically given by
![\frac{dV_0}{dV}=\frca{R}{R_1r_3}](https://tex.z-dn.net/?f=%5Cfrac%7BdV_0%7D%7BdV%7D%3D%5Cfrca%7BR%7D%7BR_1r_3%7D)
Therefore
![\triangle V_0=\triangle V*\frac{R}{R_fR}](https://tex.z-dn.net/?f=%5Ctriangle%20V_0%3D%5Ctriangle%20V%2A%5Cfrac%7BR%7D%7BR_fR%7D)
![\triangle V_0=1*\frac{8}{8*82}](https://tex.z-dn.net/?f=%5Ctriangle%20V_0%3D1%2A%5Cfrac%7B8%7D%7B8%2A82%7D)
![\triangle V_0=0.08V](https://tex.z-dn.net/?f=%5Ctriangle%20V_0%3D0.08V)
Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft
Answer:
The following statements are true:
A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction
C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface
E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.
Select ALL statements that are TRUE
B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant
D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed
Explanation:
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
![\Delta S=\dfrac{Q}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cdfrac%7BQ%7D%7BT%7D)
Now by putting the values
![20.3=\dfrac{Q}{400}](https://tex.z-dn.net/?f=20.3%3D%5Cdfrac%7BQ%7D%7B400%7D)
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ