Ask question

Ask question

All categories

3 years ago

13

"Design a sequential circuit with two T flip-flops A and B, and one input x. When x = 0, the circuit remains in the same state.

When x = 1, the circuit goes through the state transitions from 00 to 01, to 10, to 11, back to 00, and repeats. What is the boolean equation of the input of flip-flop A (TA)? Type variables and operations without blanks."

1 answer:

Harlamova29_29 [7]3 years ago

3 0

Answer:

See attached images for the diagrams and tables

You might be interested in

g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress

zvonat [6]

Answer:

a) $\mathbf{\sigma _ 1 = 4800 psi}$

$\mathbf{ \sigma _2 = 0}$

b) $\mathbf{\sigma _ 1 = 6000 psi}$

$\mathbf{ \sigma _2 = 3000 psi}$

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = 0$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 4800 psi}$

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = \frac{Pd}{4t}$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 6000 psi}$

$\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}$

$\mathbf{ \sigma _2 = 3000 psi}$

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0

4 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$

$\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}$

$\sigma_c = 81.5MPa$

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

$\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}$

Where,

$\sigma_a=$ Fatigue limit for comined alternating and mean stress

$S_e =$ Fatigue Limit

$\sigma_c=$ Mean stress (due to static load)

$\sigma_u =$ Ultimate tensile stress

$Fs =$ Security Factor

We can replace the values and assume a security factor of 1, then

$\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}$

Re-arrenge for $\sigma_a$

$\sigma_a = 254.8Mpa$

We know that the stress is representing as,

$\sigma_a = \frac{M_c}{I}$

Then,

Where $M_c$ =Max Moment

I= Intertia

The inertia for this object is

$I=\frac{\pi d^4}{64}$

Then replacing and re-arrenge for $M_c$

$M_c = \frac{\sigma_a*\pi*d^3}{32}$

$M_c = \frac{260.9*10^6*\pi*0.05^3}{32}$

$M_c = 3201.7N.m$

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0

4 years ago

Who is the mystical body of Christ

Sergeu [11.5K]

If you're talking ab jesus christ , some religions believe it was god in human form and some believe it was gods son it all depends on the religion and how it's taught

8 0

3 years ago

Read 2 more answers

What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them

OlgaM077 [116]

Yes I will answer soon

7 0

3 years ago

What type of engineering do you think would help solve this SDG???

OleMash [197]

Answer:

Explanation:

Planning engineering

4 0

3 years ago