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Drupady [299]
3 years ago
12

Both comet types come from some location near the ecliptic plane of the Solar System, but the long-period comets come from farth

er away. B. Both comet types come from random directions in a spherical distribution around the Sun, but the long-period comets come from farther away. C. Short-period comets come from a random direction in a spherical distribution around the Sun, and long-period comets come from farther away, in some location near the ecliptic plane of the Solar System. D. Short-period comets come from some location near the ecliptic plane of the Solar System, and long-period comets come from farther away, with random directions in a spherical distribution around the Sun.
Geography
1 answer:
iogann1982 [59]3 years ago
8 0

Answer:

D. Short-period comets come from some location near the ecliptic plane of the Solar System, and long-period comets come from farther away, with random directions in a spherical distribution around the Sun.

Explanation:

  • A comet is an icy rocky and small system body that passe near the sun get warmed and begins to release gases and is due to the effects of the solar radiation and have a highly eccentric elliptical orbit.
  • The long-period comets are those that have more than 200 yeas to complete and originate from the Oort Cloud.
  • And the short periods of comets are those that have less than 200 days of orbital periods as they have a low inclination.
  • The periods are less than 20 years and they are subject to a gravitational disturbance around larger bodies.
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What is the 1000 year period after the fall of rome called?
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What is the 1000 year period after the fall of Rome called?

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Calculate the amount, in grams, of an original 300-gram sample of radioactive isotope Potassium-40 remaining after
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Answer:

The current amount of the Potassium-40 sample is approximately 37.521 grams.

Explanation:

The amount of the sample of the radioactive isotope decays exponentially in time, the amount of mass of the sample as a function of time (m (t)), in grams, is described below:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

Where:

m_{o} - Initial mass, in grams.

t - Time, in years.

\tau - Time constant, in years.

The time constant can be found from half life (t_{1/2}), in years, described in statement:

\tau = \frac{t_{1/2}}{\ln 2} (2)

If we know that m_{o} = 300\,g, t = 3.9\times 10^{9}\,yr and t_{1/2} = 1.3\times 10^{9}\,yr, then the current amount of the sample is:

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\tau \approx 1.876\times 10^{9}\,yr

m = (300\,g)\cdot e^{-\frac{3.9\times 10^{9}\,yr}{1.876\times 10^{9}\,yr} }

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Explanation:

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2 years ago
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