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ozzi
2 years ago
12

What is the reason for the development of new construction of materials for human use? (Select all that apply.)

Engineering
1 answer:
salantis [7]2 years ago
7 0

Answer:

All 4 could be justified.

Explanation:

They all represent ultimate improvement.

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-Electronic control modules can easily evaluate the voltage and current levels of circuits to which they are connected and deter
erma4kov [3.2K]

Answer:

multiplexing

Explanation:

3 0
2 years ago
A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str
Charra [1.4K]

Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy =  370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = ln\frac{1}{1-0.2} = 0.223

so

yield strength after cold is

yield strength = K \varepsilon ^n

yield strength =  600*0.223^{0.21)

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = \frac{\sigma u}{1-W}

ultimate strength = \frac{440}{1-0.2}

ultimate strength after cold work = 550 MPa

8 0
3 years ago
An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0
3 years ago
To prevent drainage of the transmission fluid from the converter when the
-BARSIC- [3]
Answer
D I think
Explanation
4 0
2 years ago
What is the measurement below?
Bess [88]

Explanation:

इसिसिसिसैस्स्स्स्स्स्स्स्स्स्सूस्सोस्स्स्स्स्स

8 0
2 years ago
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