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iren [92.7K]
4 years ago
9

What is differences Between hard shoulder & soft shoulder in civil Engineerin?

Engineering
1 answer:
r-ruslan [8.4K]4 years ago
5 0

Answer:

<em><u>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.'</u></em>

<em><u>Mark</u></em><em><u> </u></em><em><u>as</u></em><em><u> brilliant</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> </u></em>

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The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

6 0
3 years ago
What do you guys like in engineering
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Answer:

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Explanation:

3 0
3 years ago
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The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th
Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}

Given that t= 200 days

R(200)=e^{-\left(\dfrac{200}{300}\right)^2}

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

0.95=e^{-\left(\dfrac{t}{300}\right)^2}

by solving above equation t=67.74 days

When R=90 %

0.90=e^{-\left(\dfrac{t}{300}\right)^2}

by solving above equation t=97.2 days

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3 years ago
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Gre4nikov [31]

Answer:

Option A,  World War II

Explanation:

During the period of industrial revolution around 1915-25, the chemical engineering has taken a new shape. During this period (i.e around the world war I), there was rise in demand for  liquid fuels, synthetic fertilizer, and other chemical products. This lead to development of chemistry centre in Germany . There was rise in use of synthetics fibres and polymers. World war II saw the growth of catalytic cracking, fluidized beds, synthetic rubber, pharmaceuticals production, oil & oil products, etc. and because of rising chemical demand, chemical engineering took a new shape during this period

Hence, option A is the right answer

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