Answer:
![\mathbf{Q_p =682 \ \ ft^3/s}](https://tex.z-dn.net/?f=%5Cmathbf%7BQ_p%20%20%3D682%20%5C%20%5C%20%20ft%5E3%2Fs%7D)
Explanation:
Given that:
Area = 475 acres
The length of the channel (L) = 6870 feet
The average water shield slope (S) = 100 feet/mile
Since; 1 mile = 5280 feet
Burst duration D = 15 min
∴
100 feet/mile = 100/5280
The average water shield slope (S) = 5/264
Using hydrograph method:
The time of concentration ![t_c = 0.0078L^{0.77} S^{-0.385}](https://tex.z-dn.net/?f=t_c%20%3D%200.0078L%5E%7B0.77%7D%20S%5E%7B-0.385%7D)
where;
L = 6870
S = 5/264
![t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}](https://tex.z-dn.net/?f=t_c%20%3D%200.0078%286870%29%5E%7B0.77%7D%20%28%5Cdfrac%7B5%7D%7B264%7D%29%5E%7B-0.385%7D)
min
Since 60 min = 1 hour
32.34 min will be (32.34*1)/60
= 0.539 hour
Lag time ![T_l = 0.67\times t_c](https://tex.z-dn.net/?f=T_l%20%3D%200.67%5Ctimes%20t_c)
![T_l = 0.67\times 32.34](https://tex.z-dn.net/?f=T_l%20%3D%200.67%5Ctimes%2032.34)
![T_l = 21.6678\ min](https://tex.z-dn.net/?f=T_l%20%3D%2021.6678%5C%20min)
The time to peak i.e
![T_p = \dfrac{D}{2}+ T_L \\ \\ T_p = \dfrac{15}{2}+ 21.6678 \\ \\ T_p = 29.168 \ min](https://tex.z-dn.net/?f=T_p%20%3D%20%5Cdfrac%7BD%7D%7B2%7D%2B%20T_L%20%5C%5C%20%5C%5C%20%20T_p%20%3D%20%5Cdfrac%7B15%7D%7B2%7D%2B%2021.6678%20%5C%5C%20%5C%5C%20%20T_p%20%3D%2029.168%20%5C%20min)
![T_r = \dfrac{T_p}{5.5} \\ \\ T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min](https://tex.z-dn.net/?f=T_r%20%3D%20%5Cdfrac%7BT_p%7D%7B5.5%7D%20%5C%5C%20%5C%5C%20%20T_r%20%3D%20%5Cdfrac%7B29.1678%7D%7B5.5%7D%20%5C%5C%20%5C%5C%20T_r%20%3D%205.30%20%5C%20min)
Since D = 15 min is not equal to
, then we hydrograph apart from
duration lag time.
Then;
![T_p \ ' = T_p + \dfrac{D-t_r}{4} \\ \\ T_p \ ' = 29.168 + \dfrac{15-5.30}{4} \\ \\ T_p \ ' = 31.593](https://tex.z-dn.net/?f=T_p%20%5C%20%27%20%3D%20T_p%20%2B%20%5Cdfrac%7BD-t_r%7D%7B4%7D%20%5C%5C%20%5C%5C%20T_p%20%5C%20%27%20%3D%2029.168%20%2B%20%5Cdfrac%7B15-5.30%7D%7B4%7D%20%5C%5C%20%5C%5C%20T_p%20%5C%20%27%20%3D%2031.593)
Now, we need to determine the peak discharge
by using the formula:
![Q_p = \dfrac{484 \times A}{T_p \ '}](https://tex.z-dn.net/?f=Q_p%20%20%3D%20%5Cdfrac%7B484%20%5Ctimes%20A%7D%7BT_p%20%5C%20%27%7D)
where
484 = peak factor
Recall that A = 475 acres, to miles, we have:
A = 0.7422 mile²
![T_p \ ' = 31.593/60](https://tex.z-dn.net/?f=T_p%20%5C%20%27%20%3D%2031.593%2F60)
∴
![Q_p = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}](https://tex.z-dn.net/?f=Q_p%20%20%3D%20%5Cdfrac%7B484%20%5Ctimes%200.7422%7D%7B%5Cdfrac%7B31.593%7D%7B60%7D%7D)
![\mathbf{Q_p =682 \ \ ft^3/s}](https://tex.z-dn.net/?f=%5Cmathbf%7BQ_p%20%20%3D682%20%5C%20%5C%20%20ft%5E3%2Fs%7D)
Answer:
d
Explanation:
baka lang nmn po.........
They are the last ones to die or get extinct
Answer:
Cylindrical map projections
Explanation:
Cylindrical map projections are used for portraying the Earth. Cylindrical map projections are rectangles, but are called cylindrical because they can be rolled up and their edges mapped in a tube, or cylinder. They have straight coordinate lines with horizontal parallels crossing meridians at right angles. All meridians are equally spaced and the scale is consistent along each parallel. The only factor that distinguishes different cylindrical map projections from one another is the scale used when spacing the parallel lines on the map.
Cylindrical map projections are great for comparing latitudes to each other and are useful for teaching and visualizing the world as a whole, by determining continents, languages, etc but really aren’t the most accurate way of visualizing how the world really looks in its entirety.