Answer:
The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Explanation:
Length of curve is given as
![L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft](https://tex.z-dn.net/?f=L%3D2%28PVT-PVI%29%5C%5CL%3D2%28242%2B30-240%2B00%29%5C%5CL%3D2%28230%29%5C%5CL%3D460%20ft)
is given as
![G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%](https://tex.z-dn.net/?f=G_2%3D%5Cfrac%7BE_%7BPVT%7D-E_%7BPVI%7D%7D%7B0.5L%7D%5C%5CG_2%3D%5Cfrac%7B127.5-122%7D%7B0.5%2A460%7D%5C%5CG_2%3D0.025%3D2.5%20%5C%25)
The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as
![A=\frac{L}{K}\\A=\frac{460}{115}\\A=4](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BL%7D%7BK%7D%5C%5CA%3D%5Cfrac%7B460%7D%7B115%7D%5C%5CA%3D4)
A is given as
![-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%](https://tex.z-dn.net/?f=-G_1%3DA-G_2%5C%5C-G_1%3D4.0-2.5%5C%5C-G_1%3D1.5%5C%5CG_1%3D-1.5%5C%25)
With initial grade, the elevation of PVC is
![E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\](https://tex.z-dn.net/?f=E_%7BPVC%7D%3DE_%7BPVI%7D%2BG_1%28L%2F2%29%5C%5CE_%7BPVC%7D%3D122%2B1.5%25%28460%2F2%29%5C%5CE_%7BPVC%7D%3D125.45%20ft%5C%5C)
The station is given as
![St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\](https://tex.z-dn.net/?f=St_%7BPVC%7D%3DSt_%7BPVI%7D-%28L%2F2%29%5C%5CSt_%7BPVC%7D%3D24000-%28230%29%5C%5CSt_%7BPVC%7D%3D237%2B70%5C%5C)
Low point is given as
![x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft](https://tex.z-dn.net/?f=x%3DK%20%5Ctimes%20%7CG_1%7C%5C%5Cx%3D115%20%5Ctimes%201.5%5C%5Cx%3D172.5%20ft)
The station of low point is given as
![St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\](https://tex.z-dn.net/?f=St_%7Blow%7D%3DSt_%7BPVC%7D-%28x%29%5C%5CSt_%7Blow%7D%3D23770%2B%28172.5%29%5C%5CSt_%7Blow%7D%3D239%2B42.5%20ft%5C%5C)
The elevation is given as
![E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft](https://tex.z-dn.net/?f=E_%7Blow%7D%3D%5Cfrac%7BG_2-G_1%7D%7B2L%7D%20x%5E2%2BG_1x%2BE_%7BPVC%7D%5C%5CE_%7Blow%7D%3D%5Cfrac%7B2.5-%28-1.5%29%7D%7B2%2A460%7D%20%281.72%29%5E2%2B%28-1.5%29%2A%281.72%29%2B125.45%5C%5CE_%7Blow%7D%3D124.16%20ft)
So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Answer:
Use GitHub or stackoverflow for this answer
Explanation:
It helps with programming a lot
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
![dQ = m \times c_p \times (T_2 -T_1)](https://tex.z-dn.net/?f=dQ%20%3D%20m%20%5Ctimes%20c_p%20%5Ctimes%20%28T_2%20-T_1%29)
For ideal gas,
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
Answer:
D Reflection
Explanation:
It reflects the light from the sun. Then when the earth gets in the way, it casts a shadow causing crescent moons.
Answer:
52, 50, 54, 54, 56
Explanation:
The "stem" in this scenario is the tens digit of the number. Each "leaf" is the ones digit of a distinct number with the given tens digit.
5 | 20446 represents the numbers 52, 50, 54, 54, 56