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pshichka [43]
3 years ago
13

In the construction of a timber-framed house, select and justify three timber based

Engineering
1 answer:
kvasek [131]3 years ago
3 0

Timber frame construction uses timber studs and rails, together with a structural sheathing board, to form a structural frame that transmits all vertical and horizontal loads to the foundations.

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Click this link to view O*NET’s Wages and Employment section for Film and Video Editors.
vova2212 [387]

Answer:

much faster than average

Explanation:

did it on edge (2022-2032)

6 0
2 years ago
When designing a car that runs on wind or Air car . can you tell me the details for the following points Compressed Air Engine:
BabaBlast [244]

Answer:

a)

The crack and connecting rod is used in the design of car.This mechanism is known as slider -crank mechanism.

Components:

1.Inlet tube

2. Wheel

3. Exhaust

4. Engine

5.Air tank

6.Pressure gauge

7.Stand

8. Gate valve

b)

The efficiency of air engine is less as compare to efficiency of electric engine and this is not ecofriendly because it produce green house gases.These gases affect the environment.

c)

it can run around 722 km when it is full charge.

                                                                                                                                                     

5 0
3 years ago
The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0
2 years ago
96/64 reduced to its lowest term
Marina CMI [18]

Answer:

3/2

Explanation:

8 0
2 years ago
Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,
Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take p_o = 880 kg/m³ and p_{glycerol = 1260 kg/m³    

 

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

Given that;

Inlet velocity of Glycerin, V_A = 6 m/s

Inlet velocity of oil, V_B = 3 m/s  

Density velocity of glycerin, p_{glycerol = 1260 kg/m³

Density velocity of glycerin, Take p_o = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, d_A = 100 mm = 0.1 m

Diameter of oil pipe, d_B = 80 mm = 0.08 m

Diameter of outlet pipe d_C = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

Q_A + Q_B = Q_C

A_Av_A + A_Bv_B = A_Cv_C

π/4 × (d_A)²v_A + π/4 × (d_B )²v_B = π/4 × (d_C)²v_C

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²v_C

0.04712 + 0.0150796 = 0.0113097v_C

0.0621996 = 0.0113097v_C

v_C = 0.0621996 / 0.0113097

v_C  = 5.5 m/s

Now we apply the mass flow rate condition

m_A + m_B = m_C

p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C  

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p  

72.6412 = 0.06220335p    

p = 72.6412 / 0.06220335

p =  1167.8 kg/m³  

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

4 0
3 years ago
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