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pav-90 [236]
3 years ago
14

You have just finished your OST takeoffs for a single-story home and found 175 LF of interior walls with 2x6 studs 14" OC. The h

ome is 8 ft tall and has two top plates and one bottom plate. What is the total BF for the interior walls?
Engineering
1 answer:
zimovet [89]3 years ago
7 0

Answer:

Total BF for the interior wall is 7.50BD

Explanation:

Given Data:

· Size of stud = 2” x 6”

· Height of Wall = 8 ft

· Top plates = 2

· Bottom Plate = 1

BF stands for board feet in lumber/wood terminology. It is the unit of volume.

1 BF (Board feet) = 1 ft x 1 ft x 1 inch

Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.

Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft

Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD

Total BF for the interior wall is 7.50BD

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Answer: N has to be lesser than or equal to 1666.

Explanation:

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Cost of parts N in gate array = $3N + $20000

Cost of parts N in standard cell = $1N + $100000

So,

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(cost of FPGA lesser than that of gate array)

 Also. 15N < 1N + 100000  lets say this is equation 2  

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Now

From equation 1

12N < 20000

N < 1666.67

From equation 2

14N < 100000

N < 7142.85

AT the same time, Both conditions must hold true

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N has to be lesser than or equal to 1666.

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3 years ago
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2 years ago
Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i
Brilliant_brown [7]

Answer:

\dot Q_{in} = 372.239\,MW

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

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h_{out} = w_{in}+h_{in}

h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}

h_{out} = 203.81\,\frac{kJ}{kg}

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

h_{out} = 2685.4\,\frac{kJ}{kg}

The rate of heat transfer in the boiler is:

\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)

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3 0
3 years ago
Read 2 more answers
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

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Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

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Answer:

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