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Iteru [2.4K]
3 years ago
8

Which actions would the maintenance and operations crews carry out as a building is completed and preparing to open to the publi

c? Select all that apply.
make any last minute repairs


install new equipment and furnishings


troubleshoot solutions to any lingering problems or malfunctions


determine which new residents or company employees will occupy which offices or residences
Engineering
2 answers:
pochemuha3 years ago
5 0

Answer:

1. make any last minute repairs

2. install new equipment and furnishings

3. troubleshoot solutions to any lingering problems or malfunctions

Explanation:

Lena [83]3 years ago
4 0

Answer:

1. make any last minute repairs

2. install new equipment and furnishings

3. troubleshoot solutions to any lingering problems or malfunctions

Explanation:

Given that the role or functions of maintenance and operations crews in building construction are to ensure that the building is maintained to the required standard of use.

This involves making sure that all repairs in a building are done. The equipment is properly installed and solved any form of equipment that malfunctions.

Hence, in this case, the correct answer is:

1. make any last-minute repairs

2. install new equipment and furnishings

3. troubleshoot solutions to any lingering problems or malfunctions

You might be interested in
The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr
djyliett [7]

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

The strength of the steel with the grais size of 0.004 mm is 347.39 MPa.

6 0
3 years ago
The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo
Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)

consider the equation of bernoulli at the top of the pool

P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)

where P_1=P_0 atm pressure

At the top of the pool v_1=0m/s, substitute in V_1 in equation (2)

P_0+\rho gh_1 =constant ...(3)

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)

where P_2=P_0 atm pressure and h_2=0m

P_0+\rho v_1^2 =constant ...(6)

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

        P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)    

Hence velocity is v_2=(\sqrt{2gh_1})m/s

d.) consider (7)

v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)

This is the norminal value of velocity  

e.) consider the equation of flow rate interval of v and t

flow(t)=\frac{dv}{dt}(m^3/s) hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2

6 0
3 years ago
Discuss the impact of the changing urban center. Include the impacts on political, economic, and social roles and opportunities.
KengaRu [80]

Answer:

The 21st century world have been earmarked with great influx of people to the urban centre,the notion of gender equality and female education have also made most traditional roles in the family changing.Before now,wives we're known for their full independence on their husband who is considered the bread winner.

Inspite of the growth of of the urban centre the availability of resources have dwindled,resulting in the surge of unemployment in many urban centre,the political entity of the society which is the government have serious challenging in managing the various threat posed by overpopulation, unemployment results in the decrease of standard of living of person and family,to cater for this family have to change their roles,wives now work to support the husband.

Explanation:

6 0
3 years ago
How many ase certifications are there for automotive technicians?
romanna [79]

Answer:

There are 50 ASE certification tests, covering almost every imaginable aspect of the automotive repair and service industry.

Explanation:

yww <33

5 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is
Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0
3 years ago
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