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2 years ago

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

Engineering

1 answer:

Valentin [98]2 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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What is differences Between hard shoulder & soft shoulder in civil Engineerin?

r-ruslan [8.4K]

Answer:

The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'

Mark as brilliant........... 

5 0

3 years ago

A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition Cβ = 87

we use here lever rule that is

Wα = Wβ ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0

3 years ago

The distribution of actual weights of 8‑ounce wedges of cheddar cheese produced at a dairy is Normal, with mean 8.1 ounces and s

s344n2d4d5 [400]

Answer:

sampling distribution

Explanation:

Sampling distribution is distribution of multiple samples' satistics of a population.

3 0

3 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$