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wariber [46]
3 years ago
11

Recall that photosynthetic rates remain relatively constant in regions near the equator. Imagine that tropical environments pers

isted throughout Earth's northern and southern hemispheres (i.e., Earth's entire climate mirrored that near the equator). If Keeling had collected his atmospheric CO2 data on such an Earth, what would you expect the Keeling Curve to look like?
a) a straight line without a slope (atmospheric CO2 levels would have remained constant over time)
b) a sinusoidal curve sloping downward (atmospheric CO2 levels would fluctuate seasonally, but would have decreased over time)
c) a straight line sloping downward (atmospheric CO2 levels would not seasonally oscillate, but would have decreased over time)
d) a sinusoidal curve sloping upward (atmospheric CO2 levels would fluctuate seasonally, but would increase over time)
e) a straight line sloping upward (atmospheric CO2 levels would not seasonally oscillate, but would have increased over time)
Physics
1 answer:
solniwko [45]3 years ago
8 0

Answer:E.

a straight line sloping upward (atmospheric CO2 levels would not seasonally oscillate, but would have increased over time)

Explanation: Photosynthesis is the process through which green plants and certain other organisms make their food using carbon dioxide and water in the he presence of Sunlight.

During photosynthesis, energy conversion from light energy into chemical energy takes place.

Photosynthesis is one of the major processes that helps to eliminate and reduce the quantity of carbon dioxide in the atmosphere,it also serve to make oxygen present in the atmosphere.

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Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p
Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

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v = [tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}[/tex]

v = 387 x 10⁻⁵ m/s

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Time taken to fall a distance of 100 m

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