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3 years ago

14

Earth whose in situ weight is 105lb/cf and whose compacted weight is 122 lb/cf is placed in a fill at the rate of 260 cy/hr, mea

sured as compacted earth. The thickness of the compacted layers is 7 in. A roller having a drum width of 50 in operates at a speed of 2 mph. Assume a 55-min hour operation efficiency. Determine the number of rollers required to provide the necessary compaction if four passes are specified for each layer of earth.

1 answer:

Stells [14]3 years ago

6 0

Answer:

Number of rollers required to complete the compaction are 2

Explanation:

The solution is given in the attachments.

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A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

$E=\frac {\sigma}{\epsilon}$

Making $\sigma$ the subject then

$\sigma=E\epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain

Since strain is given as 0.025% of the length then strain is $\frac {0.025}{100}=0.00025$

Now substituting E for $10\times 10^{6} psi$ then

$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

As already mentioned, $2t=d_o - d_i hence t=0.5(d_o - d_i)$

$t=0.5(2-1.785)=0.1075 in$

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4 years ago

Which of the following statements about glycogen metabolism is FALSE?

Ratling [72]

Answer:

Glycogen is the primary energy source for muscle and liver cells.

Explanation:

Glycogen is a readily mobilized storage form of glucose. It is a very large, branched polymer of glucose residues that can be broken down to yield glucose molecules when energy is needed. Most of the glucose residues in glycogen are linked by α-1,4-glycosidic bonds. Branches at about every tenth residue are created by α-1,6-glycosidic bonds.

Glycogen is not as reduced as fatty acids are and consequently not as energy rich. Why do animals store any energy as glycogen? Why not convert all excess fuel into fatty acids? Glycogen is an important fuel reserve for several reasons. The controlled breakdown of glycogen and release of glucose increase the amount of glucose that is available between meals. Hence, glycogen serves as a buffer to maintain blood-glucose levels. Glycogen's role in maintaining blood-glucose levels is especially important because glucose is virtually the only fuel used by the brain, except during prolonged starvation. Moreover, the glucose from glycogen is readily mobilized and is therefore a good source of energy for sudden, strenuous activity. Unlike fatty acids, the released glucose can provide energy in the absence of oxygen and can thus supply energy for anaerobic activity.

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3 years ago

A cylindrical tank is required to contain a gage pressure 520 kPa . The tank is to be made of A516 grade 60 steel with a maximum

enot [183]

Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

Maximum allowable normal stress ,σ= 150

d= 2.6 m

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The normal stress for pressure vessel given as

$\sigma=\dfrac{Pd}{2t}$ ( hoop stress)

We always take maximum stress for safe design.

$\sigma=\dfrac{Pd}{2t}$

Now by putting the values

$150\times 1000=\dfrac{520\times 2.6}{2t}$

t= 4.5 mm

So the minimum thickness, t, of the wall is 4.5 mm

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MA_775_DIABLO [31]

Answer:

generate

define

present

evaluate

develop

construct and test

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Answer:

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Explanation:

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