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3 years ago

2. What effect did the geography of ancient Greece have on its early development?

Geography

1 answer:

Alik [6]3 years ago

7 0

Answer:

The geography of ancient Greece determined and restricted its development to only certain areas, such as sea trade.

Explanation:

Ancient Greece, or rather the Hellenic city-states, are considered as one of the most developed civilizations of their time, though there can be made a solid case against it. Anyhow, in the early stages of the development of these city-states, they were very restricted in their development because of geographic factors.

The climate was of the Mediterranean type, which is a good climate, but problem was that the topography was rugged and mountainous, so agriculture was very limited to only a handful of crops and herding. Because all of the city-states had access to the sea, they focused on developing trade through it, with goods such as grapes, wine, olives, olives, and olive oil the most traded. They needed good wood for building ships, and they didn't really have any of that type. This made them heavily dependent on the Macedonians who had an abundance of wood, and considering that there was open hate between the two, it was not an ideal situation.

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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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