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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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Impact of electricity in the society

Travka [436]

Answer:

<h2>Electricity has many uses in our day to day life. It is used for lighting rooms, working fans and domestic appliances like using electric stoves, A/C and more. All these provide comfort to people. In factories, large machines are worked with the help of electricity.</h2>

5 0

2 years ago

If the result of your calculation of a quantity has si units kg•m^2/s^2•C, that quantity could be

Novosadov [1.4K]

It would be Joules.
Workdone is measured in Joules.
Workdone = Force * distance
Force = mass * acceleration
= kg * ms⁻²
= kgms⁻²

Distance = m

So, Force * distance
kgms⁻² * m

Apply laws of indices that says
x² * x³ = x²⁺³ = x⁵

Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2

4 0

3 years ago

Read 2 more answers

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0

3 years ago

One problem with digital data is that it can be vulnerable to hackers.

hoa [83]

Answer: c. A person who gains unauthorized access to digital data

Explanation:

5 0

2 years ago

The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg

kumpel [21]

The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

$\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\$