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2 years ago

8

Were well-educated members of various Catholic religious orders.

1 answer:

frosja888 [35]2 years ago

8 0

Yes they are very well educated

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Describe the terrain geography of the city of Stockholm, the country of Sweden, and transportation challenges that result becaus

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Sweden is a northern European country located on the Scandinavian Peninsula. Its topography consists mainly of flat or gently rolling lowlands but there are mountains in its western areas near Norway. Its highest point, Kebnekaise at 6,926 feet (2,111 m) is located here

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2 years ago

What are the implications of greater urbanization in Eastern Asia

andriy [413]

The implications of greater urbanization in Eastern Asia are the following;

<span>- </span>The government must invest more on the following;

<span>- </span>Housing

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<span>- </span>Water

This are required in order to have a greater urbanization in the Eastern Asia.

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2 years ago

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3 de las principales resoluciones de la Asamblea Deliberante de 1825

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SDFGHJKL;][POIUYTR

Explanation:GHJ

DFGHJKL;IUYTREHJK

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2 years ago

Chapter 3: Models of the Earth

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2 years ago

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

$T_p \ ' = 31.593/60$

∴

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

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

7 0

2 years ago