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Pachacha [2.7K]
3 years ago
15

What event triggered the dramatic increase in atmospheric carbon dioxide concentration seen over the last couple of centuries? a

) the Tunguska meteoric impact b) the Industrial Revolution c) the invention of the automobile d) the invention of CFCs e) the volcanic eruption in Iceland in 1783
Physics
1 answer:
Mice21 [21]3 years ago
3 0

Answer:

a) the Tunguska meteoric impact

Explanation:

The Tunguska Event, sometimes known as the Tungus Meteorite is thought to have resulted from an asteroid or comet entering the earth's atmosphere and exploding. The event released as much energy as fifteen one-megaton atomic bombs. As well as blasting an enormous amount of dust into the atmosphere, felling 60 million trees over an area of more than 2000 square kilometres. Shaidurov suggests that this explosion would have caused "considerable stirring of the high layers of atmosphere and change its structure." Such meteoric disruption was the trigger for the subsequent rise in global temperatures

According to Vladimir Shaidurov of the Russian Academy of Sciences, the apparent rise in average global temperature recorded by scientists over the last hundred years or so could be due to atmospheric changes that are not connected to human emissions of carbon dioxide from the burning of natural gas and oil.

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A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).
just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles (\vec r_{cm}), measured in meters, is defined by this weighted average:

\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}} (1)

Where:

m_{i} - Mass of the i-th particle, measured in kilograms.

\vec r_{i} - Location of the i-th particle with respect to origin, measured in meters.

If we know that \vec r_{cm} = (-0.500\,m,-0.700\,m), m_{1} = 1\,kg, \vec r_{1} = (-1.20\,m, 0.500\,m), m_{2} = 4.50\,kg, \vec r_{2} = (0.600\,m, -0.750\,m) and m_{3} = 4\,kg, then the coordinates of the third particle are:

(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}

(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})

(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)

(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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