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Natasha_Volkova [10]
4 years ago
7

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

r atmosphere is ozone, \(\rm O_3\). In particular, ozone absorbs radiation with frequencies around 9.38×1014 \({\rm \rm Hz}\) . What is the wavelength \(\texttip{\lambda }{lambda}\) of the radiation absorbed by ozone?
Physics
1 answer:
Sergeeva-Olga [200]4 years ago
5 0

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x  10⁻⁹ m

λ = 319.83 nm

Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm

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Why is the overall charge of the atom neutral or zero?
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Answer:

B

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For two objects with masses 100 kg and 500 kg to be 20 meters apart , find the gravity between them
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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

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Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

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2 years ago
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