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Ivanshal [37]
2 years ago
13

Build a 3-input XOR gate using a 3x8 decoder and an Or gate.please Quickly i have final​

Engineering
1 answer:
lana [24]2 years ago
7 0

Answer:

Take a 3 to 8 decoder with active low outputs

Assuming you are familiar with the functioning of decoders,

The three inputs of decoder of course are the first, second and the carry bit which you feed to the subtractor.

Next we examine the truth table of the full subtractor i formatted in the picture.

Then write the minterms for the difference output and borrow output from the given truth table picture I have mentioned before!!

Explanation:

hopi it to help you!!

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It has been estimated that 139.2x10^6 m^2 of rainforest is destroyed each day. assume that the initial area of tropical rainfore
Dmitry [639]

Answer:

A. 6.96 x 10^-6 /day

B. 22.466 x 10^12 m^2

C. 9.1125 x 10^14 kg of CO2

Explanation:

A. Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Initial area of the rainforest = 20 x 10^12 m^2

Therefore to calculate exponential rate in 1/day,

Rate of rainforest destruction/ initial area of rainforest

= 139.2 x 10^6/20 x 10^12

= 6.96 x 10^-6 /day

B. Rainforest left in 2015 using the rate in A.

2015 - 1975 = 40 years

(40 * 365 )days + 10 days (leap years)

= 14610 days

Area of rainforest in 1975 = 24.5 x 10^12m^2

Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Area of rainforest in 2015 = 14610 * 139.2 x 10^6

= 2.034 x 10^12 m^2

Area left = area of rainforest in 1975 - area of rainforest destroyed in 40years

= 24.5 x 10^12 - 2.034 x 10^12

= 22.466 x 10^12 m^2

C. How much CO2 will be removed in 2025

Recall: Photosynthesis is the process of plants taking in CO2 and water to give glucose and O2.

So CO2 removed is the same as rainforest removed so we use the rate of rainforest removed in a day

Area of rainforest in 1975 = 24.5 x 10^12 m^2

Area of rainforest removed in 2025 = 18262 days * 139.2 x 10^6

= 2.54 x 10^12 m^2

Area of rainforest removed between 1975 - 2025 = 24.5 x 10^12 - 2.54 x 10^12

= 21.958 x 10^12 mC2 of rainforest removed

CO2 = 0.83kg/m^2.year

CO2 removed between 1975 - 2025 = 0.83 * 21.958 x 10^12 * 50 years

= 9.1125 x 10^14 kg of CO2 was removed between 1975 - 2025

6 0
3 years ago
For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.
rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN,  Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

n_{p} = \frac{S_{p}A_{t}  }{CP_{max}+ F_{i}  }  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = \frac{380*20.1}{0.277*7500*5728.5} = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

n_{L} =  \frac{S_{p}A_{t}-F_{i}   }{C(P_{max} )}  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : n_{L} = 0.9191n_{L}  = 0.9191

<u>C)  Determine the factor of safety based on joint separation</u>

n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence n_{0} = 1.056

<u>D)  Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

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There 11,238,513 different Powerball
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••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

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