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3 years ago

Which area is the most common source region

2 answers:

6 0

Answer – 3. Central Canada
The area that is the most common source region for cold, dry air masses that move over New York State is central Canada. These cold, dry air masses are called Continental polar (cP) or continental arctic (cA) air masses. Continental polar air masses are one of the five air masses that affect the United States during the course of a typical year. <span>
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liq [111]3 years ago

4 0

Answer:

Central Canada

Explanation:

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In nuclear fission and fusion reactions mass is converted into energy. The average person in US uses around 100,000,000 J of ene

daser333 [38]

Answer:

a) The yearly energy consumption for a US population of 300 million is $3\times 10^{16} J/year$

b) The energy that would be released if a 60 kg person were converted entirely into energy is $5.4\times 10^{18}$ Joules.

c) 180 years would this amount of energy support a population of 300 million.

Explanation:

a) Average energy consumed by single person of US = 100,000,000 J/year

Then 300 million US citizen will consume:

300 million = 300 × 1,000,000 = $3\times 10^8$

The yearly energy consumption for a US population :

$100,000,000 J/year\times 3\times 10^8=3\times 10^{16} J/year$

The yearly energy consumption for a US population of 300 million is $3\times 10^{16} J/year$

b) $E = m\times c^2$

E = Energy from converted mass of m

c = speed of light

Given mass of person = m = 60 kg

$E=60 kg\times (3\times 10^8 m/s)^2 = 5.4\times 10^{18} J$

c) Energy calculated in part (b) = $E=5.4\times 10^{18} J$

The yearly energy consumption for a US population of 300 million in an year = $3\times 10^{16} J/year$

Let the that would be supported by $5.4\times 10^{18}$ Joules of energy be x.

$x\times 3\times 10^{16} J/year=5.4\times 10^{18} J$

$x=\frac{5.4\times 10^{18} J}{3\times 10^{16} J/year}=180 years$

180 years would this amount of energy support a population of 300 million.

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