Answer:
The theoretical maximum specific gravity at 6.5% binder content is 2.44.
Explanation:
Given the specific gravity at 5.0 % binder content 2.495
Therefore
95 % mix + 5 % binder gives S.G. = 2.495
Where the binder is S.G. = 1, Therefore
Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495
(95 +5)/(Vx +5) = 2.495
2.495 × (Vx + 5) = 100
Vx =35.08 to 95
Or density of mix = Mx/Vx = 95/35.08 = 2.7081
Therefore when we have 6.5 % binder content, we get
Per 100 mass unit
93.5 Mass unit of Mx has a volume of
Mass/Density = 93.5/2.7081 = 34.526 volume units
Therefore we have
At 6.5 % binder content.
(100 mass unit)/(34.526 + 6.5) = 2.44
The theoretical maximum specific gravity at 6.5% binder content = 2.44.
Answer:
If you are destroying the environment or habitats or trees. Also if people live nearby. Also if its legal.
Explanation:
Answer:
Explanation:
Complete question:
Fill in the blanks
One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that time ,___1______the agency ends. In addition, the parties can specify that the agency is for a particular____2______ . Once that is achieved, the agency ends. Alternatively, the parties can include a specific event as a trigger for termination; once that event,_____3______ the agency ends. The parties can terminate an agency relationship prior to any of the preceding events by ______4_________agreement, or revocation_____5______ by individual party.
Answer
1) lapses
(2) purpose
(3) occurs / begins
(4) mutual
(5) either
Answer:
can't understand the writting
Answer:
a. 30°
b. 0.9MPa
Explanation:
The slip will occur along that direction for which the Schmid factor is maximum. The three possible slip directions are mentioned as 30°, 48°, 78°
The cosines for the possible λ values are given as
For 30°, cos 30 = 0.867
For 48°, cos 48 = 0.67
For 78°, cos 78 = 0.21
Among the three-calculated cosine values, the largest cos(λ) gives the favored slip direction
The maximum value of Schmid factor is 0.87. Thus, the most favored slip direction is 30° with the tensile axis.
The plastic deformation begins at a tensile stress of 2.5MPa. Also, the value of the angle between the slip plane normal and the tensile axis is mentioned as 65°
Thus, calculate the value of critical resolved shear stress for zinc:
From the expression for Schmid’s law:
τ = σ*cos(Φ)*cos(λ)
Substituting 2.5MPa for σ, 30° for λ and 65° for Φ
We obtain The critical resolved shear stress for zinc, τ = 0.9 MPa
