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2 years ago

Write the output expression for a NOR gate with inputs , , and .

Engineering

1 answer:

mezya [45]2 years ago

4 0

Answer:

Logic NOR Gate Equivalent

The Boolean expression for a logic NOR gate is denoted by a plus sign, ( + ) with a line or Overline, ( ‾‾ ) over the expression to signify the NOT or logical negation of the NOR gate giving us the Boolean expression of: A+B = Q.

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A refrigerator operating on the Carnot cycle is used to make ice. Water freezing at 32oF is the cold reservoir. Heat is rejected

Mnenie [13.5K]

Answer:

Explanation:

From the given information:

Water freezing temp. $T_L = 32 ^0 \ F$

Heat rejected temp $T_H = 72 ^0 \ F$

Recall that:

The coefficient of performance is:

$COP_{ref} = \dfrac{T_L}{T_H - T_L} \\ \\ = \dfrac{32+460}{(72 +460) -(32+460)} \\ \\ =\dfrac{492}{532 -492} \\ \\ = \dfrac{492}{40} \\ \\ COP_{ref} = 12.3$

Again:

The efficiency given by COP can be represented by:

$COP = \dfrac{Q_L}{W} \\ \\ W = \dfrac{Q_L}{COP} \\ \\ W = \dfrac{2000 \ lbm \times 144 \ Btu/lbm}{12.3} \\ \\ W = 23414.63 \ Btu$

Finally; the power input in an hour can be determined by using the formula:

$Power= \dfrac{W}{t} \\ \\ Power = \dfrac{23414.63 \ Btu}{ 1 \ hr} \\ \\ Power = \dfrac{23414.634 \times 1055.056 \ J}{1 \times 3600} \\ \\ Power = 6.86 \ kW$

In hp; since 1 kW = 1.34102 hp

6.86kW will be = (6.86 × 1.34102) hp

= 9.199 hp

7 0

3 years ago

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Answer:

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Explanation:

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Answer:

forwarder

Explanation:

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3 years ago

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

kondaur [170]

Answer:

Detailed solution is given in the attached diagram

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3 years ago

The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j

Anna [14]

Answer:

the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

Explanation:

Given that;

volume of cut = 25,100 m³

Volume of dry soil fill = 23,300 m³

Weight of the soil will be;

⇒ 93% × 18.3 kN/m³ × 23,300 m³

= 0.93 × 426390 kN 3

= 396,542.7 kN

Optimum moisture content = 12.9 %

Required amount of moisture = (12.9 - 8.3)% = 4.6 %

So,

Weight of water required = 4.6% × 396,542.7 = 18241 kN

Volume of water required = 18241 / 9.81 = 1859 m³

Volume of water required = 1859 kL

Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

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3 years ago

Write the output expression for a NOR gate with inputs , , and .​

Write the output expression for a NOR gate with inputs , , and .