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3 years ago

What is the phase of the Moon if it . . .

Physics

1 answer:

anzhelika [568]3 years ago

8 0

Answer:

Waxing Gibbous

Third quarter

Waning Gibbous

Explanation:

If moon rises at 3:00 pm then the phase of the moon will be "Waxing Gibbous".

This is because, the moon is actually not fully illuminated but has achieved more than half of its full illumination.

If the moon is highest in the sky at sunrise then the phase of the Moon will be the "Third quarter"

This is because of the fact that at this position moon will rise at midnight, thus it will be at the highest point at the time of the sunrise.

If the moon sets at 10:00 am then the phase of the Moon is "Waning Gibbous"

This is because of the fact that at this position the Moon is moving towards becoming new Moon but at the same time, the moon is illuminated more than its half illumination.

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11 months ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

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En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

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