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GalinKa [24]
4 years ago
6

If a soil has e=0.72, moisture content = 12% and Gs=2.72, find the weight of water in KN/m3 to be added to make the soil saturat

ed.
Engineering
1 answer:
Doss [256]4 years ago
7 0

Answer:

453 KN/m^3 of water should be added make the soil saturated.

Explanation:

As we know

Gs * w = S*e

S = Gs * w/e

S = 2.72 * 0.12/0.72 = 0.453

The amount of water to be added

0.453 *1000/10 = 453 KN/m^3

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If pure oxygen is fed in excess by 25%, what would the fractional conversion of methane be for the final concentration of CO2 in
inessss [21]

Answer:the fractional conversion of methane is 12.5%

Explanation:The reaction represent the combustion of methane to produce Co2 and steam.

CH4 +2O2_CO2 + 2H2O

From gay lussac law of proportionality

1mol of CH4 requires 2mol of Oxygen to produce 1 mol of CO2 and 2mol of H2O

So from the combining ratio,25% of O2 will fractional produce 25×1/2% of CH4.

While 12.5% of CO2 and 25% of steam is also produced .so in essence 2.5% of CO2 was lost in the reaction.

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4 years ago
Explain the difference in the heat transfer modes of conduction and convection.
LiRa [457]

Explanation:

Conduction is the heat transfer through a stationary matter by the physical contact.  

For example, the heat transferred between electric burner of stove and bottom of pan is transferred by the process of conduction.

Convection is heat transfer by macroscopic movement of the fluid. The particles of the fluid carry the current within themselves.  

For example, the water is the pot is warmed overall by heat transferred by the process of convection.

4 0
3 years ago
Can you tell me important facts about Peggy A. Whitson?
Nat2105 [25]

Answer:

- She’s A Biochemist

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6 0
3 years ago
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An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into
olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}

For the 9 mm:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391

For the 0.44 caliber:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460

The ratio is 460 / 391 = 1.176

6 0
3 years ago
The following are related to wastewater treatment:
Paraphin [41]

Answer:

A. - Primary wastewater treatment: the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater.

- Secondary wastewater treatment: treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process.

-Tertiary wastewater treatment: Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending on the type of water contamination that must be removed.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C. - Activated Sludge, Rotating biological contractors.

Explanation:

A. Primary wastewater treatment refers to sedimentation, the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater. The main goal of primary treatment is to remove organic and inorganic solids that settles by sedimentation, and by act of skimming removed materials end up floating.

Secondary wastewater treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process. The goal of secondary treatment is the further treatment of the effluent from primary treatment in order to get rid of the residual organics and suspended solids.

Advanced wastewater is quite in between. Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending upon the type of water contamination that must be removed. Advanced treatment processes can be combined with primary or secondary treatment or used instead of secondary treatment.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C.

-Activated Sludge: The activated sludge process is a kind of wastewater treatment process for treating sewage/industrial wastewaters by the use of aeration and a biological floc which consists of bacteria and protozoa.

- Rotating biological contactor: Rotating biological contactors (RBCs) are fixed-film reactors that have similarities with biofilters in the sense that organisms are attached to support media.

Activated sludge is a suspended process, which is when the biomass is mixed with the sewage while in rotating biological contractor, the biomass grows on the media and then, the sewage passes over the surface.

6 0
4 years ago
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