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vredina [299]
3 years ago
7

Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have fou

r arithmetic operations which depend on the values of the selection variables as shown below. V1=0011, V0=0101 and output functions are as follows;
1- A+9's complement of B
2- A+B
3- A+10's complement of B
4- A+1 (add 1 to A)
(You can see question number 3 in the attached file)
Engineering
1 answer:
Bingel [31]3 years ago
4 0
Ucsaaaaauxx627384772938282’cc ed un e uff ridicolizzarla +golfista
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The autorotation spin characteristics of a straight-wing aircraft are induced by Group of answer choices
NemiM [27]

Answer:

More Drag on the down going wing and More Lift on the up going wing

Explanation:

The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.

Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.

As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.

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3 years ago
malott, m. e. (2003). paradox of organizational change: engineering organizations with behavioral systems analysis.
mylen [45]

Answer:

Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.

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1 year ago
Which of these is not a type of socket
kotegsom [21]
You didn’t put a picture
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2 years ago
HAPPINESS DISCUSSION
RideAnS [48]

Answer:

uh because life sucks o_<

8 0
2 years ago
Read 2 more answers
A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the
muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, \L_e = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

P =\frac{\pi^2EI}{L_e^2}

where, I is the moment of inertia

on substituting the respective values, we get

10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}

or

I = 13.13 in⁴

also for circular cross-section

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

thus,

13.13 = \frac{\pi}{64}\times d^4

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0
3 years ago
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