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3 years ago

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

Engineering

1 answer:

mezya [45]3 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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Oxygen enters an insulated 14.2-cm-diameter pipe with a velocity of 60 m/s. At the pipe entrance, the oxygen is at 240 kPa and 2

MatroZZZ [7]

Answer:

Entropy generation==0.12 KW/K

Explanation:

$s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$

$s_2-s_1=0.891\ln \frac{291}{293}-0.2598\ln \frac{200}{240}$

$s_2-s_1=0.0412\frac{KJ}{kg-K}$

Mass flow rate= $\rho\times\dfrac{\pi}{4}d^2V$

$\rho_1=\dfrac {P_1}{RT_1}$

$\rho_1=\dfrac {240}{0.2598\times 293}$

$\rho_1=3.51\frac{kg}{m^3}$

mass flow rate= $\rho_1A_1V_1$

So by putting the values

Mass flow rate=2.97 kg/s

So entropy generation=(2.97)(0.0412)

=0.12 KW/K

8 0

4 years ago

A 220-V electric heater has two heating coils that can be switched such that either coil can be used independently or the two ca

Nana76 [90]

Answer:

The resistances of both coils are 131.7 Ω and 29.64 Ω.

Explanation:

Since, there are two coils, they can be used independently or in series or parallel. The power is given as:

Power = P = VI

but, from Ohm's Law:

V = IR

I = V/R

therefore,

P = V²/R

R = V²/P

Hence, the resistance (R) and (P) are inversely proportional. Therefore, the maximum value of resistance will give minimum power, that is, 300 W. And the maximum resistance will be in series arrangement, as in series the total resistance gets higher than, any individual resistance.

Therefore,

Rmax = V²/Pmin = R1 + R2

R1 + R2 = (220 V)²/300 W

R1 + R2 = 161.333 Ω ______ en (1)

Similarly, the minimum resistance will give maximum power. And the minimum resistance will occur in parallel combination. Because equivalent resistance of parallel combination is less than any individual resistance.

Therefore,

(R1 R2)/(R1 + R2) = (220 V)²/2000 W

using eqn (1), we get:

(R1 R2) / 161.333 Ω = 24.2 Ω

R1 R2 = 3904.266 Ω²

R1 = 3904.266 Ω²/R2 _____ eqn (2)

Using this value of R1 in eqn (1), we get:

3904.266/R2 +R2 = 161.333

(R2)² - 161.333 R2 +3904.266 = 0

Solving this quadratic eqn we get two values of R2 as:

R2 = 131.7 Ω OR R2 = 29.64 Ω

when ,we substitute these values in eqn (1) to find R1, we get get the same two values as R2, alternatively. This means that the two coils have these resistance, and the order does not matter.

<u>Therefore, the resistance of both coils are found to be 131.7 Ω and 29.64 Ω</u>

7 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".