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White raven [17]
3 years ago
5

An interglacial is a period of time that occurs between Ice Ages. It’s marked by warmer temperatures and milder climates. During

previous interglacials, the CO2 levels in the atmosphere were about 280 parts per million (ppm). In 2013, they were approximately 400 ppm. Rounding to the nearest percent, what is the percent increase in CO2 between these two time periods?
Physics
1 answer:
Korolek [52]3 years ago
3 0

Answer:

The correct answer is - 43%.

Explanation: The increase in CO2 between these two suggested periods is approximately 43%. Even though it is a natural process that the CO2 levels vary in the atmosphere, still this is not the same case nowadays. Nowadays, or rather in the past few decades, apart from the natural increase of CO2 in the atmosphere, it has seen a much more increased levels because of the human activity. The industrial facilities and the vehicles, the cutting of the forests and burning the wood (there's both release of CO2 from the burning of the trees and loss of natural accumulator of the CO2), are just some of the more important human activities that contribute to a significant rise in the CO2 levels.

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Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
3 0
3 years ago
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konstantin123 [22]

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7 0
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We can detect that by looking at the graph, because Lilly's line
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AlekseyPX

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