Pipelines are a useful means of transporting oil because they offer low maintenance and dependable transportation for a narrow but important range of products.
<h3>What is a pipeline?</h3>
A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.
The options are attached
a. Pipelines provide jobs for consumers because of the resurgence of exploration and drilling in North America.
b. Pipelines are versatile, carrying more ton-miles than any other mode of transport over more than 2 million miles of pipeline.
c. Pipelines have more locations than water carriers.
d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.
Thus, the correct option is d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.
Learn more about Pipelines
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Answer: you’re bussy *in french accent*
Explanation:
it is very wide as you can see *also in french*
Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
Answer:
im trying to do a challenge because this kid deleted all my answers
Explanation:
Answer:
Q= - 7 KJ
Heat is rejected
Explanation:
In first process:
Process is adiabatic so heat transfer will be zero.
Q= 0 For adiabatic process
Now from first law of thermodynamics
Q = ΔU + W
ΔU is the change in internal energy
Q is the heat transfer
W is the work.
So here Q= 0
And work is transfer to the system .It means that work done on the system so it will be taken as negative.
W= - 12 KJ
Q = ΔU + W
0 = ΔU -12
ΔU = 12 KJ
It means that in first process internal energy of system increase.
In second process:
non-adiabatic process and work done by system is 5 KJ.
Q = ΔU + W
U is the point function and it does not depends on the path follows it depends only on final and initial states.
So fro second process
ΔU = - 12 KJ
Q= -12 + 5
Q= - 7 KJ
In magnitude Q= 7 KJ
It mean that heat is rejected from the system because it give negative sign.