Answer:
E-D-C-B-F-A
Explanation:
The top layer is clearly shown as not being cut by the fault line and leveled
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:
a scone cake and rock is similar because of the texture
Explanation:
I believe that would be emigrate
False
mountains form from collision of earths plates