Answer:
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provide information on Jack London’s life
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Answer:

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 
where;
L = 6870
S = 5/264

min
Since 60 min = 1 hour
32.34 min will be (32.34*1)/60
= 0.539 hour
Lag time 


The time to peak i.e


Since D = 15 min is not equal to
, then we hydrograph apart from
duration lag time.
Then;

Now, we need to determine the peak discharge
by using the formula:

where
484 = peak factor
Recall that A = 475 acres, to miles, we have:
A = 0.7422 mile²

∴

