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liberstina [14]
3 years ago
7

For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True

Engineering
1 answer:
Verdich [7]3 years ago
3 0

Answer:

B.

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table  for a NAND gate with 2 inputs is as follows:

0 0    1

0 1     1

1  0    1

1  1    0

The  truth table for an OR gate, is as follows:

0 0    0

0 1     1

1  0    1

1  1     1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0    1  1       1

0 1     1  0      1

1  0    0  1      1

1  1     0  0     0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

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Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa
DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l )  = ∝ * ΔT

                                    ( 5.4 * 10^-3 / 11.9 )  = 12.1 * 10^-6 * ( T2 - 4 )

∴  ( T2 - 4 ) =  ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

8 0
3 years ago
(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
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3 0
2 years ago
g A primary sedimentation basin is designed for an average flow of 0.3 m3/s. The TSS concentration in the influent is 240 mg/L.
yarga [219]

Answer:

(a) 0.243 m3/day

(b) 96 mg/l

(c) 0.426 m3/min

Explanation:

The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240= 9.6 mg/L

Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day

To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3

Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day

Approximately, the volume of sludge is 0.243 m3/day.

(b)

Efficiency of 60 percent is equivalent to 0.6

Efficiency=(influent concentration- flow rate)/influent concentration

0.6=(240-flow rate)/240

Flow rate= 96 mg/l

(c)

Cycle time= 0.243/0.57=0.4263157894736 m3/min

Rounded off, cycle time is 0.426 m3/day

3 0
3 years ago
What is 2+2 equal to
Anit [1.1K]
The answer is 4, lol.
4 0
3 years ago
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