Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.
Answer:
the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Explanation:
Given that;
volume of cut = 25,100 m³
Volume of dry soil fill = 23,300 m³
Weight of the soil will be;
⇒ 93% × 18.3 kN/m³ × 23,300 m³
= 0.93 × 426390 kN 3
= 396,542.7 kN
Optimum moisture content = 12.9 %
Required amount of moisture = (12.9 - 8.3)% = 4.6 %
So,
Weight of water required = 4.6% × 396,542.7 = 18241 kN
Volume of water required = 18241 / 9.81 = 1859 m³
Volume of water required = 1859 kL
Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Answer:
It will be B
Explanation:
Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.
Answer:
I do!!
Explanation:
I have to sit for 3 hours lol♀️
Answer:
The change in entropy is found to be 0.85244 KJ/k
Explanation:
In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.
P1/T1 = P2/T2
T2/T1 = P2/P1
T2/T1 = 180 KPa/120KPa
T2/T1 = 1.5
Now, the change in entropy is given as:
ΔS = m(s2 - s1)
where,
s2 = Cv ln(T2/T1)
s1 = R ln(V2/V1)
ΔS = change in entropy
m = mass of CO2 = 3.2 kg
Therefore,
ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]
Since, V1 = V2, therefore,
ΔS = mCv ln(T2/T1)
Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K
Therefore,
ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)
<u>ΔS = 0.85244 KJ/k</u>
