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anygoal [31]
3 years ago
11

Steven is starting a project that requires a specialized, experienced contractor. Which selection process is the most suitable f

or Steven's needs?
A. low-bid selection
B. qualification-based selection
C. best-value selection
D. private selection
Engineering
1 answer:
Gwar [14]3 years ago
8 0

Answer:

Option B, qualification-based selection

Explanation:

Whenever bidding is done for seeking professional services, selection of contractor should be done using qualifications-based selection.

While seeking professional service, focus is on selecting the best quality and value matter, hence the contracting agency with best experience and skill for the job is selected.

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What is the purpose of O-ring and valve seals in a cylinder head?
Andrews [41]

Answer:

its to show the shape is flat and only flat at the botom and top and you can set it up ther way and it wlll still look the same.

Explanation:

8 0
2 years ago
Problem definition
LekaFEV [45]

Answer:

ummm thats alot

Explanation:

8 0
2 years ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
Which statement about cathode ray tubes (CRTs) is true?
yuradex [85]

Answer:

CRTs consume less power than LCDs.

hope it helps (^^)

# Cary on learning

6 0
2 years ago
The National Electrical Code specifies that receptacles in certain areas of a house must have ground fault circuit interrupter p
Kaylis [27]

Answer:

b

Explanation:

the NEC has expanded the requirements for ground-fault circuit interrupters (GFCI) to protect anyone who plugs into an electrical system. Initially, it was only required for temporary wiring at construction sites and in dwelling unit bathrooms, but in recent years the Code requirements for GFCI protection have expanded to include many other areas, including commercial occupancies, fountains and swimming pools, and temporary installations, to name a few. (For a complete list of 2002 NEC references, see the sidebar below)

6 0
2 years ago
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