Answer:
2.379m
Explanation:
The width = 23m
The depth = 3m
The radius is denoted as R
The wetted area is = A
The perimeter perimeter = P
Hydraulic radius
R = A/P
The area of a rectangular channel
= Width multiplied by Depth
A = 23x3
A = 69m²
Perimeter = (2x3)+23
P = 6+23
P= 29
Hydraulic radius R = 69/29
= 2.379m
This answers the question
Thank you!
Answer: Describe the greatest power in design according to Aravena? The subject of Aravena’s recent Futuna Lecture Series in New Zealand was ‘the power of design,’ which he described as ultimately being “the power of synthesis” because, increasingly, architects are dealing with complex issues and problems.
What are the three problems with global urbanization? 1. Degraded Environmental Quality ...
2. Overcrowding ...
3. Housing Problems ...
4. Unemployment ...
5. Development of Slums...
How could you use synthesis in your life to solve problems? Hence, synthesis is often not a one-time process of solution design but is used in combination with problem understanding and solution analysis to progress towards a more complete understanding of problems and solutions over time (see Applying the Systems Approach topic for a more complete discussion of the dynamics of this aspect of the approach).
I got all three answers
Answer:
False ( B )
Explanation:
considering that the wind turbine is a horizontal axis turbine
Power generated/extracted by the turbine can be calculated as
P = n * 1/2 *<em> p</em> *Av^3
where: n = turbine efficiency
<em>p = air density </em>
<em> </em>A = πd^2 / 4
v = speed
From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine
Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.