Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1

For pipe 2

Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV




head loss (h)

Now putting the all values

So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
Answer:
Using the above algorithm matches one pair of Ghostbuster and Ghost. On each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are the same, so use the algorithm recursively on each side of the line to find pairings. The worst case is when, after each iteration, one side of the line contains no Ghostbusters or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P(
)- time algorithm.
Please add more details because I don’t know if you are using a book or passage, therefore I cannot help you unless you add more detail
Answer:
Point force (Qp) = 704 kn/m²
Explanation:
Given:
length = 19 m
Width = 0.5 m
fs = 4
Vicinity of the pile = 25
Find:
Point force (Qp)
Computation:
Point force (Qp) = fs²(l+v)
Point force (Qp) = 4²(25+19)
Point force (Qp) = 16(44)
Point force (Qp) = 704 kn/m²