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GuDViN [60]
3 years ago
12

What is "Engineering"?

Engineering
1 answer:
Liono4ka [1.6K]3 years ago
5 0
Engineering is when both the application of science and math are used to solve problems. An engineer is a person who designs, builds, or maintains engines, machines, or public works. There are six large branches of engineering such as Mechanical, Chemical, Civil, Electrical, Management, and Geotechnical, they all have hundreds of subcategories of engineering under each different branch.
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Joseph wants to practice architecture. Which compulsory assessment administered by NCARB does he need to complete?
Hunter-Best [27]

Answer:

Architect Registration Examination (ARE)

Explanation:

In this scenario, Joseph wants to practice architecture. A compulsory assessment administered by National Council of Architectural Registration Boards (NCARB) which he has to complete is an Architect Registration Examination.

An Architect Registration Examination (ARE) refers to the professional licensure examination to be taken by anyone who intends to practice architecture in the United States of America, Puerto Rico, Guam, Canada, US Virgin Islands.

The main purpose of the Architect Registration Examination is to assess an architect's knowledge, skills, and abilities on architectural best practices, procedures and services. Also, it focuses on areas relating to a building's safety, soundness and health impact on the habitants. Therefore, ARE is a prerequisite for practicing architecture across the United States of America jurisdiction.

Once an architect passes the examination, he or she would be given a license to practice architecture.

<em>Hence, for Joseph to practice architecture he must take the Architect Registration Examination developed and administered by National Council of Architectural Registration Boards (NCARB). </em>

8 0
3 years ago
Which federal agency issues the model food code?
umka2103 [35]

The FDA! (food and drug administration)

4 0
3 years ago
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Ughhh my cramps hurt sm
kozerog [31]

Answer:

Explanation:Come

tomate

5 0
3 years ago
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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
Zina [86]

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

5 0
3 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
4 years ago
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