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dolphi86 [110]
3 years ago
8

Which atmospheric gases are increasing as a result of human activities?

Physics
2 answers:
Kobotan [32]3 years ago
6 0

Answer: Obviously carbon dioxide, nitrogen argon, neon, helium, krypton.... etc etc....

Explanation: Have a great day!!!

natima [27]3 years ago
4 0

Answer:

Many of these gases occur naturally, but human activity is increasing the concentrations of some of them in the atmosphere, in particular:

carbon dioxide (CO2)

methane.

nitrous oxide.

fluorinated gases.

Explanation:

You might be interested in
Una ambulancia se aleja de una persona en línea recta a razón de 30 m/s. Si la sirena
saul85 [17]

Answer:

f_o=331.046Hz

Explanation:

Use Doppler effect equation:

The Doppler effect is a physical phenomenon where an apparent change in wave frequency is presented by a sound source with respect to its observer when that same source is in motion. The general equation is given by:

f_o=f_s\frac{v\pm v_o}{v\mp v_s} \\\\Where:\\\\f_s=Actual\hspace{3}frequency\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source

  • When the observer moves towards the source v_o is positive.
  • When the observer moves away from the source v_o is negative.
  • When the source moves towards the observer v_s is negative.
  • When the source moves away from the observer v_s is positive.

Since the problem don't give us aditional information let's assume:

v=343m/s

Which is the speed of sound in air.

And using the information provided by the problem:

v_s=30m/s\\v_o=0\\f_s=360Hz

f_o=f_s\frac{v}{v+v_s} =360*\frac{343}{343+30} =331.0455764\approx 331.046Hz

The frequency perceived by the person is 331.046Hz

<em><u>Translation:</u></em>

<em><u></u></em>

Usa la ecuación del Efecto Doppler:

El efecto Doppler es un fenómeno físico en el que una fuente de sonido presenta un cambio aparente en la frecuencia de onda con respecto a su observador cuando esa misma fuente está en movimiento. La ecuación general viene dada por:

f_o=f_s\frac{v\pm v_o}{v\mp v_s} \\\\Donde:\\\\f_s=Frecuencia\hspace{3}real\hspace{3}de\hspace{3}las\hspace{3}ondas\hspace{3}sonoras\\f_o=Frecuencia\hspace{3}observada(percibida)\\v=Velocidad\hspace{3}de\hspace{3}las\hspace{3}ondas\hspace{3}sonoras\\v_o=Velocidad\hspace{3}del\hspace{3}observador\\v_s=Velocidad\hspace{3}de\hspace{3}la\hspace{3}fuente

  • Cuando el observador se mueve hacia la fuente v_o es positivo.
  • Cuando el observador se aleja de la fuente es v_o negativo.
  • Cuando la fuente se mueve hacia el observador v_s es negativa.
  • Cuando la fuente se aleja del observador v_s es positiva.

Como el problema no nos da información adicional, supongamos que:

v=343m/s

La cuál es la velocidad del sonido en el aire.

Y utilizando la información proporcionada por el problema:

v_s=30m/s\\v_o=0\\f_s=360Hz

f_o=f_s\frac{v}{v+v_s} =360*\frac{343}{343+30} =331.0455764\approx 331.046Hz

La frecuencia percibida por la persona es 331.046Hz

5 0
3 years ago
5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.​
Aleks04 [339]

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

3 0
3 years ago
Read 2 more answers
The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0
3 years ago
What cause earthquakes?
KengaRu [80]
Question; what causes earthquakes? 
Answer; what causes earthquakes are mainly when rocks are underground and then suddenly break along a fault. 
6 0
4 years ago
Read 2 more answers
A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.
Mekhanik [1.2K]

Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

vf^2 = 1900

sqrt(vf^2) = sqrt(1900)

vf = 43.59 m/s

7 0
3 years ago
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