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Anon25 [30]
3 years ago
6

1. which part of the Earth's surface receives more direct intense solar radiation?

Geography
1 answer:
givi [52]3 years ago
5 0
The part of the Earth's surface which receives more direct intense solar radiation is the equator. Answer:A
<span>The reason is that the solar radiation depends on the latitude. When the Sun's rays strike Earth's surface near the equator, the incoming </span>solar radiation<span> is more direct. At higher</span> latitudes<span>, the angle of </span>solar radiation<span> is smaller.</span>
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Answer:

The right approach will be "Eutrophication".

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  • The mechanism through which a reservoir, body of water, including stream is eutrophic, usually resulting from either the underlying and adjacent land's inorganic and organic overflow.
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Give an example of a person in a colonized area becoming acculturated.
guapka [62]
Answer:  
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4 years ago
What does creation of metamorphic rocks have to do with plastic deformation?
pychu [463]
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8 0
4 years ago
Read 2 more answers
Please answer this fast and here only​
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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