A cylindrical wood puck (density = 0.75 g/cm3 ) is placed into a water bath. The density of water is 1.0 g/cm3. The wood puck has a diameter of 11.0 cm and a height of 3.5 cm. When the puck reaches equilibrium in the water bath, how far below the surface of the water will the bottom of the puck be?

If the puck had twice the height, how would the submerged depth change?

If the puck had twice the radius, how would the submerged depth change?

0.88cm; depth dependence. listen to your prof.

It's actually 2.63cm.

To find out how far below the surface of the water the bottom of the wood puck will be, we need to determine the buoyant force acting on the puck. The buoyant force is equal to the weight of the water displaced by the puck.

First, let's calculate the volume of the wood puck. The puck is in the shape of a cylinder, so its volume can be calculated using the formula for the volume of a cylinder: V = πr^2h, where r is the radius and h is the height.

Given that the diameter of the puck is 11.0 cm, the radius (r) is half of that, which is 5.5 cm or 0.055 meters. The height (h) is 3.5 cm or 0.035 meters.

Let's plug these values into the volume formula:
V = π(0.055^2)(0.035)

Calculating this, we find that the volume of the puck is approximately 0.4 cm^3.

Now, let's calculate the weight of the water displaced by the puck. The weight of an object can be determined using the formula: weight = mass × gravitational acceleration.

The mass of the water displaced is equal to its density multiplied by the volume of the puck. The density of water is given as 1.0 g/cm^3, so we'll convert the volume of the puck to cm^3:

0.4 cm^3 * (1 g/cm^3) = 0.4 grams

Now, let's calculate the weight of the water displaced:
weight = mass × gravitational acceleration
= (0.4 grams) × (9.8 m/s^2)
= 3.92 grams*m/s^2

The buoyant force acting on the puck is equal to the weight of the water displaced. So the buoyant force is approximately 3.92 grams*m/s^2.

Considering that the density of the wood puck is 0.75 g/cm^3, we can calculate the weight of the puck using the formula: weight = mass × gravitational acceleration.

The mass of the puck is equal to its density multiplied by its volume. We already know the density and volume of the puck. Let's calculate the mass:

mass = density × volume
= (0.75 g/cm^3) × (0.4 cm^3)
= 0.3 grams

Now, let's calculate the weight of the puck:
weight = mass × gravitational acceleration
= (0.3 grams) × (9.8 m/s^2)
= 2.94 grams*m/s^2

Since the buoyant force is greater than the weight of the puck, we can conclude that the force pushing up on the puck is stronger than the force pulling it down. This means the puck will float.

When an object floats, it displaces an amount of water equal to its own weight. Therefore, the volume of water displaced by the puck is equal to the volume of the puck, which is 0.4 cm^3.

Now, let's calculate the height of the water displaced. The height of the water displaced is the distance from the bottom of the puck to the surface of the water:

height = volume of water displaced / (πr^2)
= 0.4 cm^3 / (π(0.055^2))
≈ 1.6 cm

So, the bottom of the puck will be approximately 1.6 cm below the surface of the water when it reaches equilibrium in the water bath.