Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are 
The two long are 
The two shors are 
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
![Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3](https://tex.z-dn.net/?f=Q_c%20%3D%20%5B%282%2A122.08%29%2B%282%2A86-375%29%5D%2A14%2A0.833%5C%5CQ_c%3D4864.02ft%5E3)
That in cubic yards is equal to 
Hence, we need order 5% plus that represent with the quantity
Affects traction-Type of road surface.
Slow-moving farm equipment- Typically travels no faster than 25 mph.
Hills and Curves- Attempting to return to the roadway too quickly from here may cause loss of control, crossing into head-on traffic, and a possible head on collision.
Shoulders- Causes potential encroachment into your path of travel due to limited sight lines.
Poor nighttime lighting- Common in rural environments; causes reliance on headlights.
Answer:
A) m' = 351.49 kg/s
B) m'= 1036.91 kg/s
Explanation:
We are given;
Pressure Ratio;r_p = 12
Inlet temperature of compressor;T1 = 300 K
Inlet temperature of turbine;T3 = 1000 K
cp = 1.005 kJ/kg·K
k = 1.4
Net power output; W' = 70 MW = 70000 KW
A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;
m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))
At, 100% efficiency, plugging in the relevant values, we have;
m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))
m' = 70000/199.1508
m' = 351.49 kg/s
B) At 85% efficiency, the formula will now be;
m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))
Where η is efficiency = 0.85
Thus;
m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))
m' = 70000/(1.005*(432.09129 - 364.9189)
m'= 1036.91 kg/s
Answer:
correct answer is option C)
Explanation:
W 12 x 30 section of the beam dimension
Value of E = 29000 ks i/in²
I = 238 in⁴
weight of beams per length = 30 lbs/ft = 30 × 10⁻³ ks i/ft = 2.5 × 10⁻³ ks i/in.
distribution load = 2 k/ft = 1/6 k/in
L = 16 × 12 = 192 inch.
deflection due to distributed load = 
= 
= 0.427 in.
deflection due to self weight = 
= 
= 0.01025 inch
total deflection = 0.427 in.+ 0.01025 inch
total deflection = 0.4375 in
correct answer is option C)
