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3 years ago

What are the five types of civil engineering

2 answers:

5 0

Answer: The five types of the civil engineering projects is construction and management, geotechnical, structural, transport, water, and architecture

Explanation: Hope this helps

iogann1982 [59]3 years ago

5 0

Answer:

construction, architecture, structural, management, transport

Explanation:

hope it helps

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A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$

Putt in Eq (1):

$-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C$

Energy balance is given by:

$Q_{convection}=Q_{conduction}$

$h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L$ Eq (2)

$T(x) = a(L^2-x^2)+b$

Putting x=L

$T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC$

$\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2$

From Eq (2)

$h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C$

7 0

3 years ago

Heather is troubleshooting a computer at her worksite. She has interviewed the computer’s user and is currently trying to reprod

Luba_88 [7]

Answer:

The correct option is A

Explanation:

Heather is trying to establish a theory of probable cause. In this step of the troubleshooting process, the person troubleshooting questions the obvious and then test the theory or response given by the user to really determine the cause. Once confirmation of this theory has been achieved, the troubleshooter then tries to establish a resolution to the problem. However in the event whereby the theory is not confirmed, the troubleshooter then tries to establish a new theory.

8 0

3 years ago

The uniform slender rod has a mass m.

Nikolay [14]

Answer: 3/2mg

Explanation:

Express the moment equation about point B

MB = (M K)B

-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α

α = 3g/2L cosθ

express the force equation along n and t axes.

Ft = m (aG)t

mg cosθ – Bt = m [(3g/2L cos) (L/6)]

Bt = ¾ mg cosθ

Fn = m (aG)n

Bn -mgsinθ = m[ω^2 (L/6)]

Bn =1/6 mω^2 L + mgsinθ

Calculate the angular velocity of the rod

ω = √(3g/L sinθ)

when θ = 90°, calculate the values of Bt and Bn

Bt =3/4 mg cos90°

= 0

Bn =1/6m (3g/L)(L) + mg sin (9o°)

= 3/2mg

Hence, the reactive force at A is,

FA = √(02 +(3/2mg)^2

= 3/2 mg

The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg

6 0

3 years ago

Nanotechnology is a scientific area that deals with making or changing things that are incredibly _______________. *

Dafna1 [17]

B tiny nano means small!!

3 0

3 years ago

Read 2 more answers

An engine cylinder has a stroke of 320mm and bore 280MM. Calculate the mass of air contained in the cylinder if it is filled wit

belka [17]

Answer:

The mass of the air is 0.0243 kg.

Explanation:

Step1

Given:

Stroke of the cylinder is 320 mm.

Bore of the cylinder is 280 mm.

Pressure of the air is 101.3 kpa.

Temperature of the air is 13°C.

Step2

Calculation:

Stroke volume of the cylinder is calculated as follows:

$V=\frac{\pi }{4}d^{2}L$

$V=\frac{\pi}{4}\times(\frac{280}{1000})^{2}\times(\frac{320}{1000})$

V = 0.0197 m³.

Step3

Assume air an ideal gas with gas constant 287 j/kgK. Then apply ideal gas equation for mass of the air as follows:

PV=mRT

$m=\frac{PV}{RT}$

$m=\frac{101.3\times 1000\times 0.0197}{287\times (13+273)}$

m= 0.0243 kg.

Thus, the mass of the air is 0.0243 kg.

4 0

2 years ago