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alexandr1967 [171]
3 years ago
9

Since the engineering design process may take the engineer back to its beginning, the process is considered ________

Engineering
1 answer:
Valentin [98]3 years ago
7 0

Answer:

Cyclical

Explanation:

I looked at the next question on edgenuity and it said it in the question.

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The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
Meaning of temporary stitches​
algol13

Answer:

They are use to hold garment or fabric pieces together before pernament stitches are made

6 0
2 years ago
Read 2 more answers
A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st
Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

To know more about process click here:

brainly.com/question/29310303

#SPJ4

5 0
1 year ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
3 years ago
Before turning in a test, it would be best to
Doss [256]

Answer:

check the grammar and double check the work.Explanation:

sorry if its wrong :P

4 0
3 years ago
Read 2 more answers
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