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Snowcat [4.5K]
3 years ago
6

Zona intermedia de pozos Y Efecto de inavasion

Engineering
1 answer:
dlinn [17]3 years ago
5 0

so the answer is f because your a faliure so get out of here you fatty

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A triangular roadside channel is poorly lined with riprap. The channel has side slopes of 2:1 (H:V) and longitudinal slope of 2.
Oliga [24]

Answer:

Q = 14.578 m³/s

Explanation:

Given

We use the Manning Equation as follows

Q = (1/n)*A*(∛R²)*(√S)

where

  • Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)
  • A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)
  • S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025
  • n = Manning roughness coefficient (empirical constant), dimensionless = 0.023
  • R = hydraulic radius = A/P (m for S.I.) where :
  • A = cross-sectional area of flow as defined above,
  • P = wetted perimeter of cross-sectional flow area (m for S.I.)

we get A as follows

A = (B*h)/2

where

B = 5 m (the top width of the flowing channel)

h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m   (the deep)

A = (5 m*1.25 m/2) = 3.125 m²

then we find P

P = 2*√((B/2)²+h²)   ⇒  P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m

⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m

Substituting values into the Manning equation gives:

Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)

⇒ Q = 14.578 m³/s

8 0
3 years ago
State 3 advantages and 3 disadvantages of unit rate contract​
Zinaida [17]
6 would be the answer
7 0
3 years ago
An actual vapour compression system comprises following process represents a. 1-2 Compression process b. 2-3 Condens 1 (or heat
Gemiola [76]

Answer:

Explanation:

The deatailed diagram of VCRS is given below such

1-2=Isentropic compression in which temperature increases at constant entropy

2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)

3-4=Irreversible expansion or throttling in which enthalpy remains constant

4-1=Isobaric heat addition(Evaporation)

4 0
3 years ago
Write a grammar for a language whose sentences start with an even and non-zero number of x’s, end with an odd number of z’s, and
sveta [45]

Answer:

GRAMMAR

S -> AB

A -> xxAyy | xxyy

B -> yyBzz | yz

EXPLANATION

A is for even number of x's followed by that number of y's

B is for odd number of y's followed by that number of z's

3 0
3 years ago
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

8 0
4 years ago
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