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Andrews [41]
3 years ago
15

Show how to connect the two circuit building blocks from the previous problem to create a circuit that transforms a priority-enc

oded input into a one-hot encoded output, where the highest-priority hot signal from the input is the hot signal in the output.

Engineering
1 answer:
Pavel [41]3 years ago
8 0

Answer:

When xo =0, Do= 0, not 1

Explanation:

Use 8×3 priority encoder and 3×8 decoder

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Are the wooden pillars shown in the image below, a dead load?
Brrunno [24]

Answer:

no

Explanation:

it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls

6 0
3 years ago
A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac
WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

     = 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2

     =70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

    = (73.17 -53.3/53.3)*100

    = 37%

5 0
3 years ago
When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th
Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant

Differentiating the above equation:

\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\

\to V_A=  \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\

        = \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}

3 0
3 years ago
To revise a monthly budget, changes in which categories might need to be addressed? Check all that apply.
GREYUIT [131]

Income

amount budgeted

expenses

5 0
3 years ago
What are the optical properties of steel
dezoksy [38]

Answer:

A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.

In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.

Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.

Explanation:

3 0
3 years ago
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