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3 years ago

Design a 3-bit binary counter using S-R flip flops.

Engineering

1 answer:

Helen [10]3 years ago

3 0

Answer:

This is an asynchrnous 3-bit counter. Just note that this design is different and works differently than its synchronous counterpart. It's an easier design than its synchronous counterpart, and is not as reliable because it has delays.

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A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro

Zarrin [17]

Answer:

83.6%

Explanation:

On its current schedule, the plant can theoretically produce ...

(30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

= (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

On the proposed schedule, this production becomes ...

(30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

= (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

(2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

3 0

2 years ago

What is an isentropic process?

snow_tiger [21]

Answer: Isentropic process is the process in fluids which have a constant entropy.

Explanation: The isentropic process is considered as the ideal thermodynamical process and has both adiabatic as well as reversible processes in internal form.This process supports no transfer of heat and no transformation of matter .The entropy of the provided mass also remains unchanged or consistent.These processes are usually carried out on material on the efficient device.

5 0

3 years ago

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

3 years ago

During the recovery of a cold-worked material, which of the following statement(s) is (are) true?

Len [333]

Answer:

Some of the internal strain energy is relieved.

There is some reduction in the number of dislocations.

The electrical conductivity is recovered to its precold-worked state.

The thermal conductivity is recovered to its precold-worked state

Explanation:

The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.

4 0

3 years ago

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

padilas [110]

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

6 0

3 years ago

Read 2 more answers