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Ivenika [448]
3 years ago
14

Where else can you observe colorful light emisiions? are these emission applications related?​

Chemistry
1 answer:
aksik [14]3 years ago
4 0

Explanation:

Some examples of colorful light emissions are street lights, neon signs and, of course, fireworks. Neon signs emit light when an electric current passes through the neon gas.

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What is the tightly stretched membrane that forms the entrance to the middle ear called?
leva [86]

Answer:

I'm so happy to help you with this question!! The answer is tympanic.

Explanation:

The middle ear is separated from the outer ear by the eardrum, or tympanic  membrane, a thin piece of tissue stretched tight across the ear canal. Sounds hit the eardrum, making it move.

I really hope this helps you!! :)

7 0
3 years ago
1. If a gas exerts a pressure of 2.00 atm when 20. puffs of particles are present, how many puffs are necessary to increase the
Juliette [100K]

Si un gaz exerce une pression de 2,00 atm lorsque 20 bouffées de particules sont présentes alors pour 4,50 atm il faut 45 bouffées car 20 bouffées = 2,00 atm donc 2*20=40 +0,5 =4,50

7 0
3 years ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula
DerKrebs [107]

Answer:

1 kg

2 kg

3 kg

4 kg

Explanation:

i did the assignment

8 0
3 years ago
A 475 ml sample of a gas was collected at room temperature of 23.5 °C and a pressure of
Molodets [167]

Answer:

The volume when the conditions were altered is 0.5109 L or 510.9 mL

Explanation:

Using the general gas equation,

P1 V1 / T1 = P2 V2 / T2

where;

P1 = 756 mmHg

V1 = 475 ml = 0.475 L

T1 = 23.5°C = 23.5 + 273K = 275.5 K

P2 = 722 mm Hg

T2 = 10°C = 10 + 273 K = 283 K

V2 = ?

Rearranging to make V2 the subject of the formula, we obtain:

V2 = P1 V1 T2 / P2 T1

V2 = 756 * 0.475 * 283 / 722 * 275.5

V2 = 101, 625.3 / 198911

V2 = 0.5109 L or  510.9 mL

3 0
4 years ago
(a) Calculate the wavelength of light in vacuum that has a frequency of 5.49 ✕ 1018 Hz. 0.0546 Correct: Your answer is correct.
vfiekz [6]

Answer:

a) Wavelength of the light in vacuum = (5.46 × 10⁻¹¹) m = 0.0546 nm

b) Wavelength of the light in diamond = (2.26 × 10⁻¹¹) m = 0.0226 nm

c) Energy of one photon in vacuum = (3.638 × 10⁻¹⁵) J = (2.271 × 10⁴) eV

d) No, the energy of the photon doesn't change when it is travelling inside diamond.

Explanation:

Wavelength (λ), frequency (f) and velocity of light (v) are related as thus

v = fλ

a) v = fλ

v = velocity of light in vacuum = (3.0 × 10⁸) m/s

f = frequency of the light = (5.49 × 10¹⁸) Hz

λ = wavelength of the light = ?

λ = (v/f) = (3.0 × 10⁸) ÷ (5.49 × 10¹⁸)

= (5.46 × 10⁻¹¹) m = 0.0546 nm

b) To find the wavelength of the light in diamond, we need the refractive index of diamond. This is because light, just like all other waves, change their velocities and subsequently their wavelengths in different materials according to the refractive index of the materials.

Refractive index of diamond = 2.42 (from literature)

2.42 = (wavelength of light in vacuum) ÷ (wavelength of light in diamond)

2.42 = 0.0546 ÷ λ

λ = 0.0546 ÷ 2.42 = 0.0226 nm

c) Energy of a photon in vacuum is given as

E = hf

where E = energy in Joules = ?

h = Planck's constant = (6.626 × 10⁻³⁴) J.s

f = frequency of the light in vacuum = (5.49 × 10¹⁸) Hz

E = (6.626 × 10⁻³⁴) × (5.49 × 10¹⁸) = (3.638 × 10⁻¹⁵) J

1 eV = (1.602 × 10⁻¹⁹) J

The amount of the calculated energy in eV

= (3.638 × 10⁻¹⁵) ÷ (1.602 × 10⁻¹⁹) = (2.271 × 10⁴) eV

d) As light travels from material to material, it's velocity and wavelength changes from material to material, but the frequency of the light waves stay the same. Since the energy of the photon depends solely on this frequency, it shows that the energy of the photon stays consistent in whichever material.

Hope this Helps!!!

3 0
3 years ago
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