Questions: Um peixe se mantém na mesma profundidade na água doce ajustando a quantidade de ar em ossos porosos ou em bolsas de ar para tornar sua massa específica média igual à da água. Suponha que, com as bolsas de ar vazias, um peixe tem uma massa especifica de 1,05 g / cm^3. Qual a fração entre o volume inflado e o volume da bolsa cheia para tornar sua massa específica igual à da água?

Solution Steps

We need to find the fraction of the volume of the inflated air bladder to the total volume of the fish that will make the fish's average density equal to that of water. The fish's density without the air bladder is given as \(1.05 \, \text{g/cm}^3\), and the density of water is \(1.00 \, \text{g/cm}^3\).

Let:

• \(V_{\text{fish}}\) be the volume of the fish without the air bladder.
• \(V_{\text{air}}\) be the volume of the air bladder when inflated.
• \(\rho_{\text{fish}}\) be the density of the fish without the air bladder, which is \(1.05 \, \text{g/cm}^3\).
• \(\rho_{\text{water}}\) be the density of water, which is \(1.00 \, \text{g/cm}^3\).

The total volume of the fish with the air bladder is \(V_{\text{total}} = V_{\text{fish}} + V_{\text{air}}\).

The mass of the fish remains constant, so we can write the mass \(m\) as: \[ m = \rho_{\text{fish}} \cdot V_{\text{fish}} \]

The average density of the fish with the air bladder should be equal to the density of water: \[ \rho_{\text{avg}} = \frac{m}{V_{\text{total}}} = \rho_{\text{water}} \]

Substituting the values, we get: \[ \rho_{\text{water}} = \frac{\rho_{\text{fish}} \cdot V_{\text{fish}}}{V_{\text{fish}} + V_{\text{air}}} \]

Rearranging the equation to solve for \(V_{\text{air}}\): \[ 1.00 = \frac{1.05 \cdot V_{\text{fish}}}{V_{\text{fish}} + V_{\text{air}}} \]

Multiplying both sides by \((V_{\text{fish}} + V_{\text{air}})\): \[ V_{\text{fish}} + V_{\text{air}} = 1.05 \cdot V_{\text{fish}} \]

Subtracting \(V_{\text{fish}}\) from both sides: \[ V_{\text{air}} = 0.05 \cdot V_{\text{fish}} \]

The fraction of the volume of the air bladder to the total volume of the fish is: \[ \frac{V_{\text{air}}}{V_{\text{total}}} = \frac{0.05 \cdot V_{\text{fish}}}{V_{\text{fish}} + 0.05 \cdot V_{\text{fish}}} = \frac{0.05 \cdot V_{\text{fish}}}{1.05 \cdot V_{\text{fish}}} = \frac{0.05}{1.05} \]

Simplifying the fraction: \[ \frac{0.05}{1.05} = \frac{5}{105} = \frac{1}{21} \]

Final Answer