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umka21 [38]
1 year ago
12

Why do the lighter isotopes disappear first from the atmosphere? Where do those isotopes go?

Physics
1 answer:
denpristay [2]1 year ago
3 0

Lighter molecules move fast and escape from the upper atmosphere relatively quickly.

To find the answer, we have to know more about the lighter isotopes.

<h3>What are lighter isotopes?</h3>
  • Lighter molecules are mobile and soon leave the higher atmosphere.
  • A particular element's stable isotopes have slightly different atomic masses and quantum mechanical energies.
  • The lighter isotope of an element's chemical bonds are more easily broken than the heavier isotope's.
  • As a result, the light isotope typically benefits from chemical reactions.

Thus, we can conclude that, lighter molecules move fast and escape from the upper atmosphere relatively quickly.

Learn more about the isotopes here:

brainly.com/question/364529

#SPJ4

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

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The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

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Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

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4 0
3 years ago
Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th
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Let say for every 5 s of time interval the speed will remain constant

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so here we will have

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now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

d = 1298.1 ft

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3 years ago
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