91影视

Buoyancy is the upward force exerted on an object submerged in a fluid. Archimedes' principle states that any floating object displaces its own weight of fluid. In other words, the weight of the fluid displaced by an object is equal to the buoyant force acting on the object. Mathematically, we can write this principle as: \(F_b = 蟻_fV_fg\) where: - \(F_b\) is the buoyant force - \(蟻_f\) is the density of the fluid (in this case, water) - \(V_f\) is the volume of the fluid displaced by the object - \(g\) is the acceleration due to gravity

In the initial state, the boat with the wood is floating on the water. The combined weight of the boat and the wood is balanced by the buoyancy force acting on them. That means the volume of water displaced in this state is equal to the weight of the boat and the wood divided by the density of water times the acceleration due to gravity: \(V_{displaced1} = \frac{W_{boat} + W_{wood}}{蟻_fg}\)

After the wood is thrown into the water, the weight of the boat no longer includes the weight of the wood, and so the volume of water displaced by the boat will be less. The wood will also displace its own weight in water. Since wood floats on the water, the volume of water displaced by the wood will be equal to the volume of the submerged part of the wood, which is lower than the total volume of the wood. Therefore, the combined volume of water displaced in the final state can be written as: \(V_{displaced2} = \frac{W_{boat}}{蟻_fg} + \frac{W_{wood}}{蟻_fg}\)

Now, we need to compare the volumes of water displaced in the initial state (\(V_{displaced1}\)) and the final state (\(V_{displaced2}\)). If \(V_{displaced1}\) is greater than \(V_{displaced2}\), then the water level will go down. If \(V_{displaced1}\) is less than \(V_{displaced2}\), then the water level will go up. If \(V_{displaced1}\) is equal to \(V_{displaced2}\), then the water level will remain unchanged. We have: \(V_{displaced1} = \frac{W_{boat} + W_{wood}}{蟻_fg}\) \(V_{displaced2} = \frac{W_{boat}}{蟻_fg} + \frac{W_{wood}}{蟻_fg}\) Since the density of water and the acceleration due to gravity are the same in both equations, we can cancel them out: \(\frac{W_{boat} + W_{wood}}{g} = \frac{W_{boat}}{g} + \frac{W_{wood}}{g}\) Since the weights on both sides of the equation are the same and nonzero, we can conclude that: \(V_{displaced1} = V_{displaced2}\)

As we have found that the volumes of water displaced in both the initial and final states are the same, the water level of the lake will remain unchanged after the wood is thrown into the lake. Therefore, the correct answer is (C) Remain unchanged.