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4 years ago

Design a sequential circuit with two D flip-flops A and B, and one input x_in.

Engineering

1 answer:

DENIUS [597]4 years ago

7 0

Answer:

View Image

Explanation:

The question is basically asking you to build a 2-bit asynchronous counter.

What the counter does is it increase it's value by 01₂ every clock pulse. So at 0₂, nothing happens, but at 1₂ it'll count up by 1. It then reset to 00₂ when it overflows.

The design for it is pretty much universal so I kinda did this from memory.

a.) A count-up counter (from 00-11) is simply made by connecting Q' to D, and the output of the previous DFF to the clock of the next one.

b.) A count-down counter (from 11-00) is simply made by using the same circuit as the count-up counter, but you connect Q' to the clock instead of Q.

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2. What is the main job of a cylinder head?

Anna11 [10]

Answer:

Explanation:

The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder

3 0

3 years ago

A 750-turn solenoid, 24 cm long, has a diameter of 2.3 cm . A 19-turn coil is wound tightly around the center of the solenoid. P

oksian1 [2.3K]

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

6 0

4 years ago

Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity

Digiron [165]

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

4 0

4 years ago

A heat engine receives 1000kJ/s of heat from a high temperature source at 600◦C and rejects heat to a cold temperature sink at 2

scoray [572]

Answer:

(A) the thermal efficiency of this engine is 66.44 %

(B) the power delivered by the engine in watts is 664,400 W

Explanation:

Given;

Input Heat received by the engine, $P_{in}$ = 1000 kJ/s