Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer:
hand tracing
Explanation:
as a programmer when we pretend computer in the debugging process by the step of each statement in recording
then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method
so while recording value of each variable each step is hand tracing
Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
- if
then you have a thin walled cylinder
or using the diameter:
- if
then you have a thin walled cylinder
The complete Question is:
Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.
What is the Rayleigh number for free convection on the outer sides of the duct?
What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?
What is the Rayleigh number for free convection on the top of the duct?
What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?
What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?
What is the total heat gain to the duct per unit length, in W/m?
Answers:
- 7709251 or 7.709 ×10⁶
- 4.87
- 965073
- 5.931 W/m² K
- 2.868 W/m² K
- 69.498 W/m
Explanation:
Find the given attachments for complete explanation
