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alex41 [277]
3 years ago
11

How is ozone formed?

Physics
2 answers:
Veronika [31]3 years ago
6 0

Answer:

The Ozone is produced naturally when solar radiation collide with oxygen molecules O_{2}.

The solar radiation is high energy, as a result of the collision, oxygen molecules are separated in atoms, this process is known as Photolysis.

After the separation process, oxygen atoms are freed and continuously colliding one another, so here they form O_{3} molecules.

The major part of the Ozone in the atmosphere is formed over the equatorial zone, because there is the most intense solar energy radiation.

frosja888 [35]3 years ago
3 0
When energetic UV rays dissociates molecules of oxygen into seperate atoms. But when they collide into each other it forms OZONE.
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If element "X" is heavier than element "Y" then...
Molodets [167]
The best answer would be C. 

The mass of an element depends on the number of particles found in the nucleus of the atom. Atomic mass can be computed by adding the number of protons and the number of neutrons. Protons and neutrons are found in the nucleus of an atom. So the answer must be letter C. 
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Sweating and panting are examples of which characteristics of life?
Marianna [84]
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A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises b
Butoxors [25]

Answer:

  Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

            y_{cm} = 1 /M ∑m_{i}  y_{i}i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

            y = -75/2 = 37.5 cm

With arms down

            y_{cm} = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

            y_{cm} = 1/70 (63 y)₀ - 7 37.5)

With arms up

          y_{cm}’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

          y_{cm}’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

        y_{cm}’ - y_{cm} = 1/70 2 (7 35.5)

         Δy = y_{cm}’ - y_{cm} = 2 7 35.5 / 70

         ΔY = 7.1 cm

7 0
3 years ago
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Un Esquiador Inicia un Salto Horizontal con una velocidad Inicial de 25 m/s. La Altura al Final de la Rampa es de 80 m del punto
Evgesh-ka [11]

Responder:

Explicación:

Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;

T = 2Usin theta / 2

theta = 90 grados

U = 25 m / s

T = 25sin90 / 2 (9,8)

T = 25 / 19,62

T = 1,27 segundos

Por lo tanto, los cielos usarán 1.27 segundos en el aire.

La distancia horizontal es el rango;

Rango R = U√2H / g

R = 25√2 (80) /9,8

R = 25√160 / 9,8

R = 25 * √16,326

R = 25 * 4.04

R = 101,02 m

Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m

5 0
3 years ago
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