Answer:
1. Soil vapor extraction - 3
2. Air sparging - 1
3. bioventing - 4
4. Natural attenuation -2
Explanation:
Soil vapor extraction - Soil vapor extraction (SVE) is the method in which perforated pipes are used into the soil and air is injected through the pipes. Contamination in soil then removed through perforation in pipes and disposed into an off-gas treatment unit.
Air sparging - In air sparging process air is injected in bubble form that remediates groundwater by volatilizing contaminants and enhancing biodegradation.
Bioventing - In bioinventing process low air flow rates are used that provide enough oxygen to sustain microbial activity in the soil and remove other contamination form the soil.
Natural attenuation - Natural attenuation is a natural process of environmental remediation which reduce the toxicity, mass, or concentration of contaminants in soil without human intervention.
We study the movement of each body separately. For the mass body A: vector: T + Na + Ga + Ff = ma * a.Scalar: on the axis Ox: ma * g * sinα-T-Ff = ma * a
On the axis Oy: Na-ma * gcosα = 0=>Na=ma * g*cosα
Ff=μ*N=μ*ma * g*cosα
ma * g * sinα-T-μ*ma * g*cosα=ma*a
For mass body B: vector: Gb + T = mb * a.Scalar: by projection of the vector relation on the axis Ox:T-mb*g=mb*a
The correct answer is c because the other three is incorrect.
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Answer:
The pressure at Mombasa is 101.396 KPascal.
Explanation:
Given: Reading on barometer = 760 mm
= 0.76 m
Density of mercury = 1.36 x
kg/![m^{3}](https://tex.z-dn.net/?f=m%5E%7B3%7D)
In this case, the pressure can be expressed as:
P = σhg
Where: σ is the density of mercury, h is height of mercury in the barometer, and g is the acceleration due of gravity.
But, g = 9.81 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
So that,
P = 1.36 x
x 0.76 x 9.81
= 101396.16 Pascal
P = 101.395 KPascal
The pressure at Mombasa is 101.396 KPascal.
Answer:
"6.67 mm" is the right solution.
Explanation:
The given values are:
- L = 2.5 m
- y = .0125
- λ = 600 nm
As we know, the equation
⇒ ![\frac{y}{L} =\frac{x \lambda}{a}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7BL%7D%20%3D%5Cfrac%7Bx%20%5Clambda%7D%7Ba%7D)
On substituting the values, we get
⇒ ![\frac{.0125}{2.5}=\frac{(1)(600\times 10^{-9}) }{a}](https://tex.z-dn.net/?f=%5Cfrac%7B.0125%7D%7B2.5%7D%3D%5Cfrac%7B%281%29%28600%5Ctimes%2010%5E%7B-9%7D%29%20%7D%7Ba%7D)
On applying cross multiplication, we get
⇒ ![.0125a=2.5 (600\times 10^{-9})](https://tex.z-dn.net/?f=.0125a%3D2.5%20%28600%5Ctimes%2010%5E%7B-9%7D%29)
⇒ ![a=\frac{2.5(600\times 10^{-9})}{.0125}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2.5%28600%5Ctimes%2010%5E%7B-9%7D%29%7D%7B.0125%7D)
⇒ ![=1.2\times 10^{-4} \ m](https://tex.z-dn.net/?f=%3D1.2%5Ctimes%2010%5E%7B-4%7D%20%5C%20m)
For new distance, we have to put this value of "a" in the above equation,
⇒ ![\frac{y}{1.5} =\frac{(1)(600\times 10^{-9})}{1.2\times 10^{-4}}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B1.5%7D%20%3D%5Cfrac%7B%281%29%28600%5Ctimes%2010%5E%7B-9%7D%29%7D%7B1.2%5Ctimes%2010%5E%7B-4%7D%7D)
⇒ ![(1.2\times 10^{-4})y=1.5(600\times 10^{-9})](https://tex.z-dn.net/?f=%281.2%5Ctimes%2010%5E%7B-4%7D%29y%3D1.5%28600%5Ctimes%2010%5E%7B-9%7D%29)
⇒ ![y=\frac{1.5(600\times 10^{-9})}{1.2\times 10^{-4}}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1.5%28600%5Ctimes%2010%5E%7B-9%7D%29%7D%7B1.2%5Ctimes%2010%5E%7B-4%7D%7D)
⇒ ![=3.22\times 10^{-3} \ m](https://tex.z-dn.net/?f=%3D3.22%5Ctimes%2010%5E%7B-3%7D%20%5C%20m)
The total distance will be twice the value of "y", we get
= ![6.67\times 10^{-3} \ m](https://tex.z-dn.net/?f=6.67%5Ctimes%2010%5E%7B-3%7D%20%5C%20m)
or,
= ![6.67 \ mm](https://tex.z-dn.net/?f=6.67%20%5C%20mm)