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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

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

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A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

3 years ago

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

3 years ago

10. An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good i

creativ13 [48]

Answer:

Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.

Explanation: lol

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4 years ago

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Marianna [84]

Answer:

like a mountain place thanks #careonlearning

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3 years ago

Natalie solved the problem below incorrectly. Identify his error and justify your reasoning. −19+3x−11+2x=2 5x−19−11=2 5x−30=2 5

arsen [322]

5x-30=2
5x=2+30 (not -28) when the -30 is brought over to the RHS, 30 should be added to 2 instead of subtracted

hence, 5x=32
x = 6.4

5 0

3 years ago