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OverLord2011 [107]
3 years ago
7

A watershed is divided into two sub-basins A and B. The surface runoff from sub-basin A enters a channel at point 1 and flows to

point 2 where the runoff from sub-basin B is added to the hydrograph and the combined flow is then routed through a reservoir. a. Determine the discharge hydrograph from the reservoir assuming that the reservoir is initially empty (i.e. no baseflow). (Only plot up to hour 12). Also, provide ERH and SHG plots for each sub-basinb. What are the areas of sub-basins A and B in square miles?
Engineering
1 answer:
Dimas [21]3 years ago
8 0

Answer:

I want to answer this question but the information provided is incomplete to plot from which the areas of sub basins A and B can be calculated.

Explanation:

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Which principle of the software engineering code of ethics has gilbert violated?.
Yuki888 [10]

Answer:

Judgement

Explanation:

Gilbert is required by the Judgement Principle to "disclose those conflicts of interest that cannot reasonably be avoided or escaped." Since Gilbert professionally believes that the software meets specifications, secures documents, and satisfies user requirements, it is not clear if he violated any principle. However, he could have informed his client of his interest in the software and also presented other software packages of different companies from which the client could make its independent choice.

7 0
3 years ago
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
3 years ago
QUESTION 3
lianna [129]
D D D D D D D D D D D D D D D DdDdddddf
6 0
3 years ago
Select the correct answer.
FrozenT [24]

Answer:

B. Italy

Explanation:

Hope this helps :)

6 0
3 years ago
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What is the primary function of NCEES?
joja [24]
National Council of Examiners for engineering and surveying a nonprofit organization
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