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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 electrical loads in _____ circuits each have the same voltage drop, with equals the total applied voltage of the circuit.
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Answer:

The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Explanation:

I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.

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When an object is dropped from the top of a tall cliff, your teacher tells you to use -10 m/s2 for the approximate acceleration
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Now, since the object is dropped from the top. The displacement should be negative, because the negative represent a downward direction.

2. Answer: velocity

The truck is travelling at 30 mph to the west. From the unit mph (miles per hour), we know its a magnitude of velocity. West indicates the direction. Hence, the given quantity is a vector -velocity.

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Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi
Nina [5.8K]

Answer:

T_2 = 0.592

Explanation:

Given

T_1 = 1.48s

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}

Take LCM

\frac{1}{C_{up}} = \frac{1+1}{C}

\frac{1}{C_{up}}= \frac{2}{C}

Cross Multiply

C_{up} * 2 = C * 1

C_{up} * 2 = C

Make C_{up} the subject

C_{up} = \frac{1}{2}C

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

C_1 = 2 * C_{up}

C_1 = 2 * \frac{1}{2}C

C_1 = C

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

C_p = C+C

C_p = 2C

So, the total capacitance (C2) is:

\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}

\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}

Take LCM

\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}

\frac{1}{C_2} = \frac{5}{2C}

Inverse both sides

C_2 = \frac{2}{5}C

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

T\ \alpha\ C

Convert to equation

T\ =kC

Make k the subject

k = \frac{T}{C}

k = \frac{T_1}{C_1} = \frac{T_2}{C_2}

\frac{T_1}{C_1} = \frac{T_2}{C_2}

Make T2 the subject

T_2 = \frac{T_1 * C_2}{C_1}

Substitute values for T1, C1 and C2

T_2 = \frac{1.48 * \frac{2}{5}C}{C}

T_2 = \frac{1.48 * \frac{2}{5}}{1}

T_2 = \frac{0.592}{1}

T_2 = 0.592

Hence, the time constance of (b) is 0.592 s

8 0
2 years ago
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