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BabaBlast [244]
3 years ago
5

Why do areas of high altitude (e.g., on the Tibetan plateau) have greater than expected UV intensity and areas of constant cloud

cover (e.g., Congo Basin) have less than expected?
Geography
1 answer:
Tju [1.3M]3 years ago
3 0

Explanation:

UV (ultraviolet) levels are high in the areas of high altitudes. Tibetan plateau being the highest plateau of the world is more exposed to the UV radiations as the atmosphere gets thinner in the areas of higher altitude. So it can not filter the maximum of ultraviolet radiations. Tibetan plateau is on 4,500 meters height and the UV level increase by 10-12% upon every 1000 meter elevation.

          However in those areas having the constant cloud cover like in Congo Basin the UV level is low as the clouds scatter the most of the UV radiations and they particularly stop the UV-B radiations. Thats why the cloudy areas have less UV level then the areas having a cloudless clear sky.

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Como el ángulo 1 mide 60°, entonces, el ángulo 2 mide 120°. El ángulo 3 mide 60°, ya que, los ángulos 1 y 3 son opuestos por el vértice, por lo tanto, tienen la misma medida.

8 0
3 years ago
Write a paragraph identifying how the Neolithic Revolution led to the establishment of early villages, and give an example. Iden
cestrela7 [59]

The Neolithic revolution was the first agricultural revolution. It was a gradual change from nomadic hunting and gathering communities and bands to agriculture and settlement. This period is described as a "revolution" because it changed the way of life of communities which made the change. It occurred in different prehistoric human societies at different times. Many societies changed 9–7 thousand years ago.

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8 0
4 years ago
3. A space vehicle is traveling at 4300 km/h relative to Earth when the exhausted rocket motor (mass 4m) is disengaged and sent
kvv77 [185]

Answer: Vce= 4234.4km/hr or 1180m/s

Explanation:

Mass of rocket motor Mr= 4m

Mass of control module Mc= m

Initial velocity of spaceship relative to earth Vs= 4300km/hr

Final velocity of command module relative to the rocket motor Vrc = 82km/hr

Total mass of spaceship M= 4m+m= 5m

Using law of conservation of momentum;

MVs= MrVre + McVce eqn1

Vre = Vrc+Vce eqn2

Where Vce is velocity of control room relative to the earth and Vre is velocity of rocket motor relative to the earth

Substituting eqn2 to eqn1, we have.

MVs= Mr(Vrc+Vce) +McVce

MVs= MrVrc+MrVce+McVce

Making Vce the subject of formula.

Vce = (MVs-MrVrc)/(Mc+Mr)

Substituting the values we have,

Vce = (5m×4300km/hr - 4m×82km/hr)/ (m+4m)

Vce = 21172m/5m

Vce= 4234.4km/hr or 1180m/s

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