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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
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
![\frac{w}{Q1} = \frac{35}{100}](https://tex.z-dn.net/?f=%5Cfrac%7Bw%7D%7BQ1%7D%20%3D%20%5Cfrac%7B35%7D%7B100%7D)
W = ![1.2*\frac{35}{100}*1000kj/s](https://tex.z-dn.net/?f=1.2%2A%5Cfrac%7B35%7D%7B100%7D%2A1000kj%2Fs)
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
![\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ](https://tex.z-dn.net/?f=%5Cfrac%7B1200%7D%7B1000%7D%20%3D%5Cfrac%7BQ2%7D%7B308%7D%20%5C%5C%5C%5CQ2%20%3D%20369.6%20KJ)
<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT
![\rho_1=\dfrac{P_1}{RT_1}](https://tex.z-dn.net/?f=%5Crho_1%3D%5Cdfrac%7BP_1%7D%7BRT_1%7D)
![\rho_1=\dfrac{350}{0.287\times 450}](https://tex.z-dn.net/?f=%5Crho_1%3D%5Cdfrac%7B350%7D%7B0.287%5Ctimes%20450%7D)
ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system
![h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}](https://tex.z-dn.net/?f=h_1%2B%5Cdfrac%7BC_1%5E2%7D%7B200%7D%2BQ%3Dh_2%2B%5Cdfrac%7BC_2%5E2%7D%7B2000%7D)
For ideal gas
Δh = CpΔT
by putting the values
![1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}](https://tex.z-dn.net/?f=1.011%5Ctimes%20450%2B%5Cdfrac%7B3%5E2%7D%7B200%7D%2BQ%3D1.011%5Ctimes%20300%2B%5Cdfrac%7B460%5E2%7D%7B2000%7D)
![Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450](https://tex.z-dn.net/?f=Q%3D1.011%5Ctimes%20300%2B%5Cdfrac%7B460%5E2%7D%7B2000%7D-%5Cdfrac%7B3%5E2%7D%7B200%7D-1.011%5Ctimes%20450)
Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding
Answer:
the two defects of a simple cell are:
1. Polarization
2. Local action