A scuba tank, when fully submerged, displaces 15.7 L of seawater. The tank itself has a mass of 13.9 kg and, when "full," contains 3.20 kg of air. The density of seawater is 1025 kg/m3. Assume that only its weight and the buoyant force act on the tank.

Question

Determine the magnitude of the net force on the fully submerged tank at the end of a dive (when it no longer contains any air).

so I tried doing 13.9+3.2 * 9.8 and then subtracting it from a previous answer where I had to find the magnitude when it was full of air. i got 157N but its not right

Answer 1

To determine the magnitude of the net force on the fully submerged tank at the end of a dive, we need to consider the weight of the tank, the buoyant force on the tank, and the change in buoyant force as the air is used up.

First, let's find the weight of the tank. We know that the mass of the tank is 13.9 kg, so the weight of the tank is $13.9\text{ kg} \times 9.8 \text{ m/s}^2 = 136.1 \text{ N}$.

Next, let's find the buoyant force on the tank. We know that the tank displaces 15.7 L of seawater, and the density of seawater is 1025 kg/m3. So the buoyant force on the tank is equal to the volume of the displaced seawater times the density of the seawater: $15.7\text{ L} \times 1\text{ m}^3/1000\text{ L} \times 1025 \text{ kg/m}^3 = 16.2 \text{ N}$.

Now, let's find the change in buoyant force as the air is used up. When the tank is full, it contains 3.20 kg of air, which displaces a volume of $\frac{3.20\text{ kg}}{1.2\text{ kg/m}^3} = 2.67\text{ m}^3$ of seawater. So the change in buoyant force is equal to the volume of displaced seawater times the density of seawater: $2.67\text{ m}^3 \times 1025 \text{ kg/m}^3 = 27.4 \text{ N}$.

Finally, we can use these values to find the magnitude of the net force on the fully submerged tank at the end of a dive: $136.1 \text{ N} - 16.2 \text{ N} + 27.4 \text{ N} = 147.3 \text{ N}$. So the magnitude of the net force on the fully submerged tank at the end of a dive is 147.3 N.

Answer 2

The net force on the fully submerged tank at the end of a dive is equal to the buoyant force, which is equal to the weight of the displaced seawater. The weight of the displaced seawater can be calculated using the density of seawater and the volume of the tank.

The volume of the tank is 15.7 L, and the density of seawater is 1025 kg/m3. Therefore, the weight of the displaced seawater is 15.7 L x 1025 kg/m3 = 16,087.5 kg.

The buoyant force is equal to the weight of the displaced seawater, so the net force on the fully submerged tank at the end of a dive is 16,087.5 kg x 9.8 m/s2 = 157,717 N.