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Alexxx [7]
4 years ago
15

the ground beneath the Nelson's house is beginning to Sag and sink which of the following is a possible explanation the ground h

as been exhausted, the Nelson's have built there home on a continental plate, there is too much salt in the soil, or leaching is causing the soul to weaken
Geography
2 answers:
Mkey [24]4 years ago
7 0

Answer:

  • <u>Option:(D)-</u>

Due to the process of leaching the soil is weakened which some how causes the unfortunate events more rarely in some regions.

Explanation:

  • <u>Evaluating other options:</u>

Tectonic plates will never effect a single house, as for the ground exhaustion there is also minimal possibility for that.

  • <u>The process of Leaching:</u>

The flow of water is always uninterrupted and as it does not move all by itself as it also carry some soluble elements. Now, these water molecules come all the way from the rivers and oceans and from high hills down to lower grounds. As due that the ground is weekend and many unfortunate may be faced by the local residents.

Ivenika [448]4 years ago
3 0
There's to much salt in the soil
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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
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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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