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To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) must you produce in your lungs?

Respuesta :

Answer:

Pressure, P = 503.48 Pascals

Explanation:

Given that,

Density of lemonade, [tex]d=1040\ kg/m^3[/tex]

Maximum height, [tex]h=4.94\ cm=0.0494\ m[/tex]

We need to find the minimum gauge pressure produced in your lungs. The pressure exerted is given by :

[tex]P=d\times g\times h[/tex]

[tex]P=1040\times 9.8\times 0.0494[/tex]              

P = 503.48 Pascals    

So, the minimum gauge pressure is 503.48 pascals. Hence, this is the required solution.                      

onyebuchinnaji

The minimum gauge pressure must you produce in your lungs is 0.005 atm.

The given parameters;

  • density of the lemonade, ρ = 1040 kg/m³
  • maximum height of the gauge, h = 4.94 cm

The minimum gauge pressure you must produce in your lung is calculated as follows;

P = ρgh

P = (1040 x 9.8 x 0.0494)

P = 503.49 Pa

The pressure in atmosphere is calculated as follows;

1 atm -------

1.01 x 10⁵ Pa ------- 1 atm

503.49 Pa ---------- ?

[tex]= \frac{503.49}{1.01 \times 10^5} = 0.005 \ atm[/tex]

Thus, the minimum gauge pressure must you produce in your lungs is 0.005 atm.

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