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3 years ago

The element in a fluorescent lightbulb that absorbs UV light and releases visible light energy is ____?

2 answers:

8 0

You're answer will be a phosphor
The element in a florescent lightbulb that absorbs UV light and releases visible light energy is --a phosphor--.

Mnenie [13.5K]3 years ago

3 0

So in a fluorescent light bulb the mercury atoms release the UV light. This UV light is absorbed by the phosphorous powder coating inside and this releases the visible light.
So your answer is phosphorous powder.

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STatiana [176]

15 degrees because a glass of water won't do anything to a bath tub of 15 degree water

4 0

3 years ago

What's the difference between meteoroids, meteorites, and meteors?

SVEN [57.7K]

You're talking about a grain of sand or a stone or a rock that's drifting in space, and then the Earth happens to get in the way, so the stone falls down to Earth, and it makes a bright streak of light while it's falling through the atmosphere and burning up from the friction.

-- While it's drifting in space, it's a meteoroid.

-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a meteor.

-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a meteorite.

4 0

3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

What is a materials ability to be dissolved in a solvent

weeeeeb [17]

Answer: Solubility

Solubility is the ability to be dissolved. Saturation is when the concentration is too high(more than solubility) that when you add another material it won't dissolve. Solute is the material that dissolved. Solvent is the material that used for dissolving

5 0

3 years ago

iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig

Alika [10]

Answer:

453 gm

Explanation:

Immersed objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm in air

8 0

2 years ago