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Radda [10]
3 years ago
6

A hypothetical accumulator processor uses one of the following 16-bit instruction formats, depending on the instruction.(a) (10

points.) Draw a block diagram of this CPU, showing the sequencer, ALU, memory, and control and data registers, and use arrows to indicate how the various parts interact with each other. Assume that instructions and data are accessed using an MAR and MDR register. (This will be the Von Neumann architecture diagram of this processor, much like the one we created in class.)(b) (20 points.) Give a control unit design for this processor using D-flip-flops similar to the one we designed in the lectures for Vesp 1.0. The design should include fetch, decode and execute steps in as much detail as possible.(c) (20 points) Give a datapath design, also using D-flip-flops for the MDR register of this processor, clearly describing all the transfers into the MDR with the help of a typical flip-flop as we did in class by making sure that you completely specify both the data and clock inputs for that flip-flop.

Engineering
1 answer:
Jlenok [28]3 years ago
7 0

Answer:

Every part has been answered with complete detail. See the pictures.

Explanation:

See the pictures for detailed answer.

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The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

y is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

Specific gravity of soil solids

G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765

Saturated Unit Weight

y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

7 0
3 years ago
Which of the followong parts does not rotate during starter operation? A. Commutator segments B. Armature windings c. Field wind
photoshop1234 [79]

Answer: B

Explanation: unless newer models added wingding to code inside fused computer...wingdings on a window ...not a motor

8 0
3 years ago
A coal-burning power plant generates electrical power at a rate of 650 megawatts (MW), or 6.50 × 108 J/s. The plant has an overa
Vinvika [58]

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce  6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x  6.50 × 10⁸ J/year

          Energy produce in one year = 204984 x 10¹² J/year

          Energy produce in one year =20.49 x 10¹⁶ J/year

7 0
3 years ago
Why is data driven analytics of interest to companies?
Svetlanka [38]

Answer:

Advantages of data analysis

Ability to make faster and more informed business decisions, backed by facts. Helps companies identify performance issues that require action. ... can be seen visually, allowing for faster and better decisions.

7 0
3 years ago
On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?.
pochemuha
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