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2 years ago

For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True

Engineering

1 answer:

Verdich [7]2 years ago

3 0

Answer:

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table for a NAND gate with 2 inputs is as follows:

0 0 1

0 1 1

1 0 1

1 1 0

The truth table for an OR gate, is as follows:

0 0 0

0 1 1

1 0 1

1 1 1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

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An electric power plant uses solid waste for fuel in the production of electricity. The cost Y in dollars per hour to produce el

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Answer:

15.64 MW

Explanation:

The computation of value of X that gives maximum profit is shown below:-

Profit = Revenue - Cost

= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2

= 14.7x - .47x^2 - 12

After solving the above equation we will get maximum differentiate for profit that is

14.7 - 0.94x = 0

So,

x = 15.64 MW

Therefore for computing the value of X that gives maximum profit we simply solve the above equation.

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3 years ago

Only put coolant into your radiator when the engine is _____.

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Only put ciilant into ur radiator when the engine is cool (D)

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2 years ago

A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the

Kay [80]

Answer:

a) V = 0.354

b) G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴ = 159.155 MPA

E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= - <u>( 19.9837 - 20 /20)</u>

2.33 × 10⁻³

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

= 68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

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3 years ago

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

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3 years ago

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lesya692 [45]

Answer:

True

Explanation:

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2 years ago