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dangina [55]
3 years ago
12

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor ack

nowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.
Engineering
1 answer:
Elza [17]3 years ago
3 0

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

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Answer:

Resistance, R = 2.04 Ohms.

Explanation:

Given the following data;

Current = 7 amps

Circuit voltage = 120 Volts

Power = 100 Watts

To find the resistance;

First of all, we would calculate the potential difference (voltage level) used by the radio.

Power = current * voltage

100 = 7 * voltage

Voltage = 100/7

Voltage = 14.29 Volts

Next, we solve for the resistance of the radio;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Substituting into the formula, we have;

14.29 = 7*R

R = \frac {14.29}{7}

Resistance, R = 2.04 Ohms.

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3 years ago
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm
igor_vitrenko [27]

Answer:

velocity and pressure in a 2.6-cm:

P2 = 2.53x10^5Pa, v2 = 1.18m/s

Explanation:

Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A

Area = πr^2

From the question:

v1 = 0.5m/s

d1 = 4cm = 0.04m

r1 = d1/2 = 0.04/2 = 0.02m

Since water was pumped from basement, h1 = 0m

P1 = 3.03x10^5 Pa

A1 = π×0.02×0.02

A1 = 0.0004πm^2

v2 = unknown

d2 = 2.6cm = 0.026m

r2 = d2/2 = 0.026/2 = 0.013m

h2 = 5m

P2 = unknown

A2 = π×0.013×0.013

A2 = 0.000169πm^2

Using continuity equation:

A1v1 = A2v2

0.0004π * 0.5 = 0.000169π * v2

v2 = (0.0004π * 0.5)/(0.000169π)

v2 = 1.18m/s

Applying a Bernoulli principle

P + 1/2*density*v^2 + density*g*h =C

C = constant

P1 + 1/2*density*v1^2 + density*g*h1

= P2 + 1/2*density*v2^2 + density*g*h2

Let g = 9.81m/s

density of water = 1000kg/m^3

(P1-P2) = 1/2* density(v2^2 - v1^2) +(density*g*h2) - (density*g*h1)

(P1-P2) = 1/2* density(v2^2 - v1^2) +

density* g(h2-h1)

(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))

(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050

3.03x10^5 - P2 = 571.2 + 49050

3.03x10^5 - P2 = 49621.2

3.03x10^5 - 49621.2 = P2

P2 = 253378.8

P2 = 2.53x10^5Pa

P2 = 2.53x10^5Pa, v2 = 1.18m/s

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3 years ago
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The current drawn by fluorescent lighting has a high total harmonic distortion. For this case, THD is calculated to be 88%. The
Alona [7]

Answer:

displacement power factor is 0.959087

Explanation:

given data

THD = 88%

true power factor = 0.72

solution

we get here total harmonic distribution THD is express as here

THD = \sqrt{\frac{1}{g^2}-1}       ..............1

her g is distortion factor

so put here value and we will get g that is

0.88² =   \frac{1}{g^2} -1    

solve it we get

g = 0.750714

and

displacement power factor is express as

DPF = \frac{PF}{g}   .................2  

put here value and we will get

DPF = \frac{0.72}{0.750714}    

DPF  = 0.959087

3 0
3 years ago
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