A large television company (CFR) contacts a technology provider, GET, Inc., to arrange a meeting to discuss an upgrade of its ex
1 answer:
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Answer:
Option D
Explanation:
Given information
Bulk unit weight of 107.0 lb/cf
Water content of 7.3%,=0.073
Specific gravity of the soil solids is 2.62
Specifications
Dry unit weight is 113 lb/cf
Water content is 6%.
Volume of embankment is 440,000-cy
Borrow material
Embankment
Considering that the volume of embankment is inversely proportional to the dry unit weight
Therefore,
Therefore, volume of borrow material is 498594-cy
(b)
The weight of water in embankment is found by multiplying the moisture content and dry unit weight.
Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf
Since
The embankment requires water of 6.78*27*440000= 80546400 lb
Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf
Borrow material requires water of 7.27959*27*498594=97998120 lb
Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb
1 gallon is approximately hence
That's approximately 4.2 gallons
Answer:
The correct answer is option (A) 0.060 uF
Note: Kindly find an attached image of the complete question below
Sources: The complete question was well researched from Quizlet.
Explanation:
Solution
Given that:
C₁ = 0.1 μF
C₂ =0.22 μF
C₃ = 0.47 μF
In this case, C₁, C₂ and C₃ are in series
Thus,
Their equivalent becomes:
1/Ceq = (1/C₁ + 1/C₂ +1/C₃
1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]
1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]
1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)
1/Ceq =[(0.1724)/(0.01034)]
1/Ceq = [(16.67)]
1/Ceq =(1/16.67) = 0.059μf
Ceq = 0.059μf ≈ 0.060μf
Therefore the equivalent capacitance of the three series capacitors is 0.060μf
Answer: hello the diagram related to your question is missing please the third image is the missing part of the question
Fx = 977.76 Ib/ft
Explanation:
<u>Estimate the force that water exerts on the pier </u>
V = 12 ft/s
D( diameter ) = 6 ft
first express the force on the first half of the cylinder as
Fx1 = - ---------------- ( 1 )
where ; Fy = 0
Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β ) ------------- ( 2 )
Input equation (2) into equation ( 1 ) (note : assuming Po = 0 )
attached below is the remaining part of the solution
