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nika2105 [10]
3 years ago
15

At a lake location, the soil surface is 6 m under the water surface. Samples from the soil deposit underlying the lake indicate

a wet unit weight of 18.8 kN/m3. Determine the effective and total vertical stresses in the deposit at a depth of 12 m below the soil surface.
Engineering
1 answer:
777dan777 [17]3 years ago
6 0

Answer:

The effective stress is 225.6 kN/m²

The total vertical stress is 338.4 kN/m²

Explanation:

Given;

depth of soil surface = 6m

weight unit of the soil samples = 18.8 kN/m³

Stress = force/area

Stress = unit weight x depth

Effective stress at a depth of 12 m below the soil surface:

= (18.8 kN/m³) x (12 m)

= 225.6 kN/m²

Effective stress = 225.6 kN/m²

Total vertical stress at a depth of 12 m below the soil surface:

= (18.8 kN/m³) x (6 m) + (18.8 kN/m³) x (12 m)

= 112.8 kN/m² + 225.6 kN/m²

= 338.4 kN/m²

Total vertical stress = 338.4 kN/m²

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Answer:

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Explanation:

given data

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tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

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3 years ago
Do all websites use the same coding to create?
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Answer:

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Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with
Scilla [17]

Answer:

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Explanation:

Given data:

circular windmill diamter D1 = 8m

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wind speed = 8 m/s

we know that specific volume is given as

v =\frac{RT}{P}

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considering air pressure is 100 kPa and temperature 20 degree celcius

v =  \frac{0.287\times 293}{100}

v = 0.8409 m^3/ kg

from continuity equation

A_1 V_1 = A_2 V_2

\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2

D_2 = D_1 \sqrt{\frac{V_1}{V_2}}

D_2 = 8 \times \sqrt{\frac{12}{8}}

D_2 = 9.797 m

mass flow rate is given as

\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}

\dot m = 717.309 kg/s

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8 0
3 years ago
Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the
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<u>Explanation:</u>

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(5)(745.7) = 3.7285 KW power delivered

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let sunlight hours be 8 hours

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=5.62×3600 = 20232 kJ/m^{2}/day

energy input in lawn = (600) (20232) (7)

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Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

                                            = 3398.97 mJ

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Trash cans repaired  

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5 0
3 years ago
Regeneration can only increase the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than the worki
storchak [24]

Answer:

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Option: A

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6 0
3 years ago
Read 2 more answers
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