ricardoredc6
contestada

A particular deep-sea fish has a swim bladder that can hold a maximum to 0.34 L of gas. It lives at 500m depth. If
the fish's swim bladder currently holds 0.12 L of gas, how high could researchers safely raise it before rupturing
("popping") the swim bladder? (Show your work)

A particular deepsea fish has a swim bladder that can hold a maximum to 034 L of gas It lives at 500m depth If the fishs swim bladder currently holds 012 L of g class=

Respuesta :

Edufirst

Answer:

  • The researcher's can safely raise the fish before reaching the 170 m depth.

Explanation:

Assuming constant temperature, you can use Boyle's law:

         [tex]p_1V_1=p_2V_2[/tex]

Where:

  • p₁ is the pressure at 500 m depth = 50.57 atm
  • p₂ is the unknown pressure
  • V₁ = 0.12 Liter
  • V₂ is the maximum volume that the fish's swim bladder can hold = 0.34 liters

Then, determine p₂:

     [tex]p_2=p_1\times \dfrac{V_1}{V_2}[/tex]

      [tex]p_2=50.57atm\times \dfrac{0.12liter}{0.34liter}=17.85atm[/tex]

From the table, there is a linear relationship between pressure (atm) and water depth (m).

The slope of the linear function is about 0.099:

  • (50 - 40)/(100-50) = 0.099

The y-intercept is 1.00 (the point 0, 1.00).

Then, the linear function is: Pressure = 1 + 0.099×depth

Now you can find the depth at wich the pressure is 17.85 atm

  • 17.85atm = 1.00atm + 0.099×depth
  • depth = (17.85 atm - 1.00atm) / 0.099 = 170.2 m ≈ 170 m

Hence, the fish's swimm bladder will ruputure at 170m.

Otras preguntas