Answer:
the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Explanation:
Given that;
volume of cut = 25,100 m³
Volume of dry soil fill = 23,300 m³
Weight of the soil will be;
⇒ 93% × 18.3 kN/m³ × 23,300 m³
= 0.93 × 426390 kN 3
= 396,542.7 kN
Optimum moisture content = 12.9 %
Required amount of moisture = (12.9 - 8.3)% = 4.6 %
So,
Weight of water required = 4.6% × 396,542.7 = 18241 kN
Volume of water required = 18241 / 9.81 = 1859 m³
Volume of water required = 1859 kL
Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Answer:
you need more details but if you have to find the difference, its $2.00
Explanation:
8-6=2
Answer:
B. Based on the allowable shear stress, but the allowable normal stress should always be checked to be sure it is not exceeded
Explanation:
Shear stress is analyzed to determine the shear forces along the lenght of the beam. This is represented in a shear force diagram. The beam cross sectional design is determined in such a way as to minimize the shear stress. Allowable normal stress should always be checked in a structure if failure is to be prevented.
Question
The mean weight of a breed of yearling cattle is 1187 pounds. Suppose that weights of all such animals can be described by the Normal model N(1187,78).
a) How many standard deviations from the mean would a steer weighing 1000 pounds be?
b) Which would be more unusual, a steer weighing 1000 pounds, or one weighing 1250 pounds?
Answer:
a. z = -2.40
A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.
b. z = 0.81
1000 is more unusual because its contained on the extreme end from the mean
Explanation:
a.
Let weight (in pounds) of the cattle be denoted by letter x:
z = (x - u)/ σ
Where u = mean and σ = standard deviation
u = 1187
σ = 78
x = 1000
Use z score formula to standardize the value of x:
z = (1000 - 1187)/78
z = -187/78
z = -2.397436
z = -2.40 ------_ Approximated
A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.
b.
x= 1250
z= (1250 - 1187)/78
z = 63/78
z = 0.807692
z = 0.81 --------- Approximated
1000 is more unusual because its contained on the extreme end from the mean
