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Ugo [173]
2 years ago
11

The following is a chemical reaction for _______ with ozone.

Physics
2 answers:
Andreas93 [3]2 years ago
6 0

a.  chlorofluorocarbons

Explanation:

The chemical reaction describes the combination of chlorofluorocarbons with ozone.

Reaction 1: ClO + BrO   ⇒  Cl + Br + O₂

Reaction 2:  Cl + O₃ ⇒ ClO + O₂

Reaction 3: Br + O₃  ⇒  BrO + O₂

O₃ is the ozone molecule, the Cl, and Br are the chlorofluorocarbons.

  • In this reaction, ozone is converted into oxygen gas. This reaction takes place in the stratosphere.
  • This leads to the penetration and creation of a hole in the atmosphere which allows for ultraviolet rays to reach the earth surface.
  • The chlorofluorocarbons are produced on the earth surface in component gases used for making refrigerants, car sprays, air conditioners etc.
  • When the reach the ozone layer, they react with it and produce oxygen gas.

learn more:

Ozone brainly.com/question/12685482

#learnwithBrainly

choli [55]2 years ago
6 0

Answer:

a.  chlorofluorocarbons

Explanation:

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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
alisha [4.7K]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

4 0
2 years ago
An electron in a mercury atom changes from energy level b to a higher energy level when the
emmainna [20.7K]
The energy is 3.06 electronvolts,  E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J


4 0
3 years ago
A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per
GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

396.97 x  .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0
3 years ago
An object has a mass of 6kg. calculate it's gpe​
m_a_m_a [10]

Explanation:

When m=<em>mass</em>

G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>

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<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>

<em>M</em><em>g</em><em>h</em>

<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>

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7 0
2 years ago
How to derive the fourth equation of motion?
Leya [2.2K]

Answer:

To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.

Explanation:

v= u + at

s =( u + v ) t /2

s = ut + at2/2

v2 = u2 + 2as

6 0
3 years ago
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