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4 years ago

1. Under what conditions can soils be chemically stabilized?

Engineering

1 answer:

marshall27 [118]4 years ago

7 0

Answer:

All will be Explained below.

Explanation:

1) Under which Condition can a soil be chemically Stabilize.

Answer

a). Plasticity Index :A soil with a high value of plasticity Index is not good for various engineering projects. The introduction of line helps in reducing plasticity due cation exchange reaction.Pozzolanic reaction over time reduces plasticity and increase index strength due to the formation of calcium - silicate hydrate.

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An engineering firm just lost one of their larger customers. The firm president says that the solution to this problem is to fir

Shkiper50 [21]

Answer:

The engineers disagreed because their jobs were on the line

The ethical factors are:

The reason for the customer dumping the business is yet to be figured out

The need to keep cost to the lowest ebb in order to keep maintain profitability at the expense of employees' welfare

The are several ways of growing customer base which are yet to be exploited.

Explanation:

The engineers disagree because there is no direct connection between the company's loss of the customer and their proposed layoff of the engineers,at least no one strong evidence has been given by the president.

The ethical factors inherent in this case are as follows:

The reason for the customer dumping the business is yet to be figure out

The need to keep cost to the lowest ebb in order to keep maintain profitability.

The are several ways of growing customer base which are yet to be exploited.

There is a need for a fact-finding exercise to establish the main motive behind losing such customer,without which the company can run into more troubles in future,otherwise the company would keep firing its good hands each time a customer dumps it.

Also,the president had resulted into such decision in order to maintain company's margins,where then lies ethics of welfare economics?Welfare economics is about looking beyond margins and looking at issues from a wider perspective of fulfilling the needs of employees in order for them to put in their best performance,at least by granting them job safety.

The company could have also grown business by investing in new technology that sets it apart from competitors instead of just jumping into the conclusions of sacking employees in a business where the company's strength lies in quality of engineers that it has.

7 0

3 years ago

A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of

OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0

3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago

Basic concepts surrounding electrical circuitry?

Elanso [62]

Hopefully that helps you out and is this for history or science?

3 0

3 years ago

Can anyone tell me what rock this is?

Oksi-84 [34.3K]

Answer: answer is B

Explanation:

8 0

3 years ago

Read 2 more answers