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lys-0071 [83]
2 years ago
9

Vince asks her friend Maria to review the construction plan of a building. Which part of the construction plan contains abbrevia

tions and signs?
Engineering
1 answer:
kaheart [24]2 years ago
7 0

Answer:

Plan symbols /Architectural symbols

Explanation:

A blueprint drawing is a drawing used in the building and construction industry to show engineers and contractors how a building, an object or a system is to be constructed and implemented including final finishing. Graphical symbols are used in these diagrams to refer readers of the drawing to other drawings in the blueprint set. Symbols make it simple for experts that prepare the drawings and further help to ensure the file document is acceptable in size.

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Initialize the tuple team_names with the strings 'Rockets', 'Raptors', 'Warriors', and 'Celtics' (The top-4 2018 NBA teams at th
Drupady [299]

Answer:

#Initialise a tuple

team_names = ('Rockets','Raptors','Warriors','Celtics')

print(team_names[0])

print(team_names[1])

print(team_names[2])

print(team_names[3])

Explanation:

The Python code illustrates or printed out the tuple team names at the end of a season.

The code displayed is a function that will display these teams as an output from the program.

4 0
2 years ago
The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air
Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0
2 years ago
A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hou
Novosadov [1.4K]

Answer:

0.867

Explanation:

The driver population factor (f_{p})can be estimated using the equation below:

f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}

The value of the heavy vehicle factor (f_{HV}) is determined below:

The values of the E_{T} = 2 and E_{R} = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

f_{HV} = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

f_{p} = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

6 0
2 years ago
What are the three most common metals used in die casting?
Anit [1.1K]

Answer:

Aluminium,Copper,Magnesium

Explanation:

The three most common metal of die casting are as follows

1.Aluminium

2.Copper

3.Magnesium

Die casting is the process in which metal is forced in the die to produces the desired casting product.Generally two type of machines are used like cold chamber and hot chamber machining,it depends on the metals.Die casting produces simple shape of casting ,it can not use for complex casting.

3 0
2 years ago
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
77julia77 [94]

Answer:

7.65 mm

Explanation:

Stress, \sigma=\frac {F}{A} where F is the force and A is the area

Also, \sigma=E\times \frac {\triangle L}{L}

Where E is Young’s modulus, L is the length and \triangle L is the elongation

Therefore,

\frac {F}{A}= E\times \frac {\triangle L}{L}

Making A the subject of the formula then

A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}

Since A=\frac {\pi d^{2}}{4} then  

d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm

4 0
2 years ago
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