A toy boat has a volume of 0.0005m^3. It has a mass of 450 grams. How much of the boat is under fresh water?

bouyant force=.450g

volume*g*densitywater=.450g
solve for volume

percent under water= volume/.0005

2.778

To determine how much of the toy boat is under fresh water, we need to compare its volume to the volume of water displaced by the boat.

Given:
- Volume of the toy boat = 0.0005 m^3
- Mass of the toy boat = 450 grams

First, let's convert the mass of the toy boat to kilograms:
Mass (kg) = Mass (grams) / 1000
Mass (kg) = 450 g / 1000 = 0.45 kg

Next, we can use Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

The buoyant force is equal to the weight of the water displaced by the boat, which is also equal to the weight of the boat itself.

Weight = Mass × gravitational acceleration
Weight of the toy boat = Mass (kg) × 9.8 m/s^2

Now, let's calculate the weight of the toy boat:
Weight of the toy boat = 0.45 kg × 9.8 m/s^2 = 4.41 N (Newtons)

Since the weight of the toy boat is equal to the weight of the water displaced, the volume of water displaced is equal to the volume of the boat.

Therefore, the amount of the boat under fresh water is its volume, which is 0.0005 m^3.