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-Dominant- [34]
3 years ago
9

The map shows a top view of the Mississippi River. The gap between lines 1 and 2 shows how much Saint Anthony Falls receded betw

een 1680 and 1766. The gap between lines 3 and 4 shows how much the falls receded between 1856 and 1876. At what rate (on average) did the falls recede during each of these time spans? Express your answer in units of feet per year. Did the rate increase or decrease from one time span to the next? Hint: Determine how many years passed between lines 1 and 2 and between lines 3 and 4 to help you find your answers.
Geography
2 answers:
DIA [1.3K]3 years ago
3 0

Answer:

The time span between 1680 and 1766 (lines 1 and 2) is 86 years: 1766 − 1680 = 86 years. During this time, the falls receded an average of 5.8 feet per year: 500 feet ÷ 86 years ≈ 5.8 feet per year.

The time span between 1856 and 1876 (lines 3 and 4) is 20 years: 1876 − 1856 = 20 years. During this time, the falls receded an average of 25 feet per year: 500 feet ÷ 20 years = 25 feet per year.

Explanation:

plato answer ^

nydimaria [60]3 years ago
3 0

Answer:

The time span between 1680 and 1766 (lines 1 and 2) is 86 years: 1766 − 1680 = 86 years. During this time, the falls receded an average of 5.8 feet per year: 500 feet ÷ 86 years ≈ 5.8 feet per year.

The time span between 1856 and 1876 (lines 3 and 4) is 20 years: 1876 − 1856 = 20 years. During this time, the falls receded an average of 25 feet per year: 500 feet ÷ 20 years = 25 feet per year.

Explanation:

the answer is correct because i took the test

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a rural kansas watershed that is ungauged has an area of 475 acres and a main channel length of 6870 feet with an average slope
dlinn [17]

Answer:

\mathbf{Q_p  =682 \ \  ft^3/s}

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}

where;

L = 6870

S = 5/264

t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}

t_c =32.34 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

T_l = 0.67\times 32.34

T_l = 21.6678\ min

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

T_r = \dfrac{T_p}{5.5} \\ \\  T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min

Since D = 15 min is not equal to T_r, then we hydrograph apart from T_r 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

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

Q_p  = \dfrac{484 \times A}{T_p \ '}

where

484 = peak factor

Recall that A = 475 acres, to miles, we have:

A = 0.7422 mile²

T_p \ ' = 31.593/60

∴

Q_p  = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}

\mathbf{Q_p  =682 \ \  ft^3/s}

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