Answer:
Δr=20.45 %
Explanation:
Given that
Rake angle α = 15°
coefficient of friction ,μ = 0.15
The friction angle β
tanβ = μ
tanβ = 0.15
β=8.83°
2φ + β - α = 90°
φ=Shear angle
2φ + 8.833° - 15° = 90°
φ = 48.08°
Chip thickness r given as
r=0.88
New coefficient of friction ,μ' = 0.3
tanβ' = μ'
tanβ' = 0.3
β'=16.69°
2φ' + β' - α = 90°
φ'=Shear angle
2φ' + 16.69° - 25° = 90°
φ' = 49.15°
Chip thickness r' given as
r'=0.70
Percentage change
Δr=20.45 %
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Answer: Volume= 1.16 yd³
Explanation:
We are given the in-situ moisture content= 18% and moist unit weight= 105 pcf
First we find out the dry unit weight
Dry unit weight= Moist unit weight / (1+ moisture content)
Dry unit weight= 
Dry unit weight= 88.983 pcf
Now, to find the volume ( in terms of each cubic yard i.e 1 yd³ we have
Volume = Dry unit weight compacted / Dry unit weight (in-situ) * 1 yd³
Volume= 
Volume= 1.16 yd³
The correct answer would be d
The objects will not move towards or away from each other
Hope this helps
