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djyliett [7]
3 years ago
10

Which of the following represents a contrast between the government of Sparta and that of Athens?

Geography
1 answer:
murzikaleks [220]3 years ago
7 0

Answer:

C. Sparta was ruled by a military dictator; Athenian government was a democratic republic.

Explanation:

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professor190 [17]

Answer:

As explained below.

Explanation:

  • Sedimentary rocks ar he secondary rocks to form over the earth due to the competition and solidification of sediments in various layers of rocks. In geologically cross-breeding is the cross-stratification of the rocky mass layering which is aligned to the angle of the bedding plane.
  • It forms depositional of the inclined surfaces such as triples and dues. And is indicated by the depositional environment of the earth like the deltas.
  • The cross-bedding tells the geologist of the area's historical significance and the conditions in which the present landscape has evolved. Other sources being the internal sorting patterns and Geometries like the tabular horizontally extended beds and the cross-beds formed by the inclined base of the rocks.
  • <u>An example of the phenomenon is cross-bedding in the Waddens Cove formation of Nova Scotia.</u>
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Vaselesa [24]

Answer:

1) a) r_{R,I} = 1150\,m, b) r_{R,II} = 3400\,m, c) r_{R,III} = 2750\,m.

2) a) r_{R,I} = 45,2\,km, b) r_{R, II} = 28,15\,km, c) r_{R,III} = 51,8\,km.

Explanation:

The scale factor is defined by the following expression:

n = \frac{r_{S}}{r_{R}} (1)

Where:

n - Scale factor.

r_{S} - Scale distance, in centimetres.

r_{R} - Real distance, in centimetres.

1) If we know that n = \frac{1}{50000}, r_{S,I} = 2,3\,cm, r_{S,II} = 6,8\,cm and r_{S,III} = 5,5\,cm, then the actual distances in metres are:

r_{R,I} = \frac{r_{S,I}}{n}

r_{R,I} = \frac{2,3\,cm}{\frac{1}{50000} }

r_{R,I} = 115000\,cm

r_{R,I} = 1150\,m

r_{R,II} = \frac{r_{S,II}}{n}

r_{R,II} = \frac{6,8\,cm}{\frac{1}{50000} }

r_{R,II} = 340000\,cm

r_{R,II} = 3400\,m

r_{R, III} = \frac{r_{S, III}}{n}

r_{R,III} = \frac{5,5\,cm}{\frac{1}{50000} }

r_{R,III} = 275000\,cm

r_{R,III} = 2750\,m

2) If we know that n = \frac{1}{50000}, r_{S, I} = 90,4\,cm, r_{S, II} = 56,3\,cm and r_{S,III} = 103,6\,cm, then the actual distance in kilometres is:

r_{R,I} = \frac{r_{S,I}}{n}

r_{R,I} = \frac{90,4\,cm}{\frac{1}{50000} }

r_{R,I} = 4520000\,cm

r_{R,I} = 45,2\,km

r_{R,II} = \frac{r_{S,II}}{n}

r_{R,II} = \frac{56,3\,cm}{\frac{1}{50000} }

r_{R,II} = 2815000\,cm

r_{R, II} = 28,15\,km

r_{R, III} = \frac{r_{S, III}}{n}

r_{R,III} = \frac{103,6\,cm}{\frac{1}{50000} }

r_{R,III} = 5180000\,cm

r_{R,III} = 51,8\,km

4 0
3 years ago
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