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stira [4]
2 years ago
10

Explain any three ways research can facilitate the work of building technicians

Engineering
1 answer:
Shtirlitz [24]2 years ago
7 0

According to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

<h3>What are building technicians?</h3>

They are experts in construction projects, supporting the design of plans, estimating costs, planning work methods, etc.

Research can facilitate the work of building technicians in the following ways:

  • Conception, design and planning of the project.
  • Selection of materials for floor, wall and ceiling systems.
  • Carry out the programming of works, budgets and analysis of unit prices.

Therefore, we can conclude that according to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

Learn more about construction projects here: brainly.com/question/17083292

#SPJ1

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A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m K), and the wire/sheath interface i
Semmy [17]

Answer:

maximum allowable electrical power=4.51W/m

critical radius of the insulation=13mm

Explanation:

Hello!

To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)

Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and  conduction as shown in the attached image

to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient

r=\frac{k}{h} =\frac{0.13}{10}=0.013m=13mm

3 0
3 years ago
A stream of moist air flows into an air conditioner with an initial humidity ratio of 0.6 kg(vapor)/kg(dry air), and a dry air f
BaLLatris [955]

Answer:

\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}

Explanation:

The final humidity ratio is computed by the Principle of Mass Conservation:

Dry Air

\dot m_{in} = \dot m_{out}

Moist

\dot m_{in} \cdot \omega_{in} + \dot m_{w} = \dot m_{out}\cdot \omega_{out}

Then, the final humidity ratio is:

\omega_{out} = \frac{\dot m_{in}\cdot \omega_{in}+\dot m_{w}}{\dot m_{out}}

\omega_{out} = \omega_{in} + \frac{\dot m_{w}}{\dot m_{out}}

\omega_{out} = 0.6\,\frac{kg\,H_{2}O}{kg\,DA} + \frac{0.4\,\frac{kg\,H_{2}O}{s} }{1.5\,\frac{kg\,DA}{s} }

\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}

4 0
4 years ago
Who wants fight with me
vladimir1956 [14]
Yea, ‘Who wants to fight with me’
3 0
3 years ago
How do you solve this. I dont know how so I need steps if you dont mind
galben [10]

Explanation:

all I know is every number that have a bar on is equal to one

4 0
3 years ago
A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
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