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iris [78.8K]
2 years ago
15

What is the role of the architects in modern development​

Engineering
1 answer:
pshichka [43]2 years ago
7 0

Answer:

to plan, design, and oversee the construction of a building

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How to make text take shape of object in affinity designer
Alina [70]

Answer:

To fit text to a shape in Affinity Designer, make sure you have your text selected. Then, grab the Frame Text Tool and click on the shape. A blinking cursor will appear within the shape, indicating that you can begin typing. The text you type will be confined to the boundaries of the shape.

Explanation:

6 0
3 years ago
What are the searching algorithms used by search engines?
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6 0
2 years ago
For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0
3 years ago
Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces
RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

mh1 + W = mh2 + Q where W = Q.

Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0
3 years ago
A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0
Aneli [31]

Answer:

Q=339.5W

T2=805.3K

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.

3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature

4 0
3 years ago
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