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Natalija [7]
3 years ago
11

Soil is to be excavated from a borrow pit to construct a 100-m-long trapezoidal levee that is 3 meters high, 5 meters wide at th

e top, and 15 meters wide at the base. The in situ total unit weight of the soil in the borrow pit is 21.7 kN/m3 with a water content of 11.4%. If levee construction requires the soil to be placed at a void ratio of 0.47, determine the volume that should be excavated from the borrow pit.
Engineering
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

The answer is 2217.32m³

Explanation:

Given that:

V₂  = (15 + 5/2) * 3 * 100 = 3000 m³

r₁ = 21.7kN/m3

w (water) = 11.4%.

e₂ (void ratio) = 0.47

V₁ = ?

Now,

rd₁ =  r₁/1 + w₁ = 21.7/1 +(11.4/100)

= 19.47 kN/m3

Thus,

rd₂ = r₁/1 + e₂ = 21.7/1+ 0.47 = 14.76 kN/m3

Then,

V₁/V₂ = rd₁/ rd₂

V₁ = 3000 * 14.76/19.97

V₁ = 2217.32m³

Therefore, volume that should be excavated from the borrow pit is 2217.32m³

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Viefleur [7K]

Answer:

coupling is in tension

Force = -244.81 N

Explanation:

Diameter of Hose ( D1 ) = 35 mm

Diameter of nozzle ( D2 ) = 25 mm

water gage pressure in hose = 510 kPa

stream leaving the nozzle is uniform

exit speed and pressure = 32 m/s and atmospheric

<u>Determine the force transmitted by the coupling between the nozzle and hose </u>

attached below is the remaining part of the  detailed solution

Inlet velocity ( V1 ) = V2 ( D2/D1 )^2  

= 32 ( 25 / 35 )^2

= 16.33 m/s

4 0
2 years ago
what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.
Leviafan [203]

Answer:

robotic technology    

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.  

3 0
3 years ago
we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw
gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0
3 years ago
Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Nutka1998 [239]

Answer:

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given :  x + 2x + 2x = 0   for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = \sqrt{\frac{k}{m} }  = \sqrt{2}  

next we determine the damping condition using the damping formula

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0
2 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

V = \frac{\pi}{4} * 0.28^2 * 5

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

7 0
3 years ago
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