Answer:
shrinkage ratio = 1.538
Explanation:
given data
water content = 22 %
dry density γ = 82 pcf
required dry density specified γ' = 96 pcf
required to produce = 50,000 yd³ = 50000 × 27 = 1350,000 ft³
solution
we get here first volume of borrow pit that is we know that
dry density ∝ 
so 
v = 1580487.8 ft³
v = 58536.58 yd³
so here
shrinkage ratio will be as
shrinkage ratio = 
shrinkage ratio = 1.538
Answer:
Soil engineering helps in analyzing the structure and composition of the soil of the proposed construction site, thus helping in deciding whether the soil of the proposed construction site or building is worth exploiting.
Answer:
The option that identifies why the bicycle cannot yet be created as a model in the scenario is;
Suzanne forgot to include the exact units of measurement that should be used
Explanation:
The design sketch turned over to the team that will work on the prototype by Suzanne should present all aspects of the design that will enable others working on the design and that make use of the sketch to have a clear understanding of what is required of them
Given that Suzanne has included the numbers that explain the relationship between the sketch and the real world object, the scale that shows the ratios and proportions of the sketch and the actual bicycle has been provided, however, given that the the machinist still need more information, the units of the measurement indicated in the drawing was not included, therefore, the actual dimensions and size that gives the length of the parts of the sketch and of the prototype to be made cannot be determined.
Answer:
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0 it is a incompressible flow.
c) ap = (16/3) i + (32/3) j + (16/3) k
Explanation:
Given
V = xy² i − (1/3) y³ j + xy k
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0
then
∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z
⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.
c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z
⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)
⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k
⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k
At point (1, 2, 3)
⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k
⇒ ap = (16/3) i + (32/3) j + (16/3) k
