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3 years ago

How often is carbon dioxide cycled through the atmosphere

Physics

1 answer:

olga55 [171]3 years ago

5 0

Maybe around 350 years, depending on the carbon cycle and the time taken through steps.

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No links Explain why the sky is blue.

Sergio039 [100]

Answer:

because of Rayleigh scattering

Explanation:

Rayleigh scattering refers to the scattering of light off of the molecules of the air, and can be extended to scattering from particles up to about a tenth of the wavelength of the light. It is Rayleigh scattering off the molecules of the air which gives us the blue sky.

6 0

3 years ago

Read 2 more answers

Io and Europa are two of Jupiter's many moons. The mean distance of Europa from Jupiter is about twice as far as that for Io and

Anastaziya [24]

The universal gravitation law and Newton's second law allow us to find that the answer for the relation of the rotation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}}$ = 2.83

The universal gravitation law states that the force between two bodies is proportional to their masses and inversely proportional to their distance squared

$F = G \frac{Mm}{r^2}$

Where G is the universal gravitational constant (G = 6.67 10⁻¹¹ $\frac{N m^2 }{kg^2}$ ), F the force, m and m the masses of the bodies and r the distance between them

Newton's second law states that force is proportional to the mass and acceleration of bodies

F = m a

Where F is the force, m the mass and the acceleration

In this case the body is the satellites of Jupiter and the planet,

$G \frac{Mm}{r^2} = m a$

Suppose the motion of the satellites is circular, then the acceleration is centripetal

a = $\frac{v^2}{r}$ r

Where v is the speed of the satellite and r the distance to the center of the planet

we substitute

$G \frac{Mm}{r^2} = m \frac{v^2}{r} \\G \frac{M}{r} = v^2$

Since the speed is constant, we can use the uniform motion ratio

v = $\frac{\Delta x}{t}$

In the case of a complete orbit, the time is called the period.

The distance traveled is the length of the orbit circle

Δx = 2π r

We substitute

$G \frac{M}{r} = (\frac{2 \pi r}{T} )^2 \\T^2 = (\frac{4 \pi ^2}{GM}) \ r^3$

Let's write this expression for each satellite

Io satellite

Let's call the distance from Jupiter is

r = $r_{Io}$

$T_{Io}^2 = (\frac{4 \pi ^2}{GM} ) \ r_{Io}^3$ TIo² = (4pi² / GM) rIo³

Europe satellite

Distance from Jupiter is

$r_{Eu} = 2 \ r_{Io}$

We calculate

$T_{Eu} = ( \frac{4\pi ^2 }{GM} (2 \ r_{Io})^3\\T_{Eu} = ( \frac{4 \pi ^2 }{GM}) r_{Io} \ 8$

$T_{Eu}^2 = 8 T_{Io}^2$

$\frac{ T_{Eu}}{T_{Io}} = \sqrt{8} = 2.83$

In conclusion, using the universal gravitation law and Newton's second law, we find that the answer for the relationship of the relation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}} = 2.83$