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DerKrebs [107]
3 years ago
5

Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Un

der which input scenario can adder generate correct output with the minimal delay? b) Under which input scenario can adder generate correct output with the maximum delay?

Engineering
2 answers:
Bogdan [553]3 years ago
5 0

Answer:

Answer for the question:

Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Under which input scenario can adder generate correct output with the minimal delay? b) Under which input scenario can adder generate correct output with the maximum delay?

Is given in the attachment.

Explanation:

Akimi4 [234]3 years ago
3 0

Answer:

With A4 + 1F input scenario, the adder can generate correct output with the minimal delay.

With AB + 55 input scenario, the adder can gcorrect output with the maximum delay.

Explanation:

A4 + 1F can be grounded to give the least significant bits and can generate correct output with the minimal delay, which AB + 55 can be grounded to give more significant bits(a0,b0) and can generate correct output with the maximum delay.

In 8-bit ripple carry adder. eight inputs a7 to a0 and b7 to b0 are applied to each of the full adder cell and output S7 to S0 represents eight bit sum from each full adder.

The input carry of the first half adder cell must be grounded for the correct addition of least significant bits (a0, b0) otherwise it will result in erroneous output

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How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

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Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

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3 years ago
What is the instantaneous center of zero velocity? List two approaches for determining the is the instantaneous center of zero v
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From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

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Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours
svet-max [94.6K]

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

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L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

C₁₀ = (2.75 KN)(2.289)

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3 years ago
What is the best engineering job to do? Why?
allochka39001 [22]

Answer:

Any engineering job would be good YOU should be the one choosing which job.

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7 0
2 years ago
Read 2 more answers
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

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Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0
3 years ago
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