In an immersion measurement of a boy’s density, he is found to have an apparent weight of 6.86-N when completely submerged in water with lungs empty. The mass of the boy is 68.0-kg in air. The density of water is 1000-kg/m^3.

a)Determine the density of the boy;
b)Will the boy float in sea water (density of sea water is 1029-kg/m^3)? If yes, what percentage of his body volume will be above sea water?

6.86 = m g - 1000 g V

g = 9.81 m/s^2
6.86 = 9.81 (68) - 9810 V
9810 V = 660
V = 66/981 m^3
m = 68
so m/V = (68/66)981 = 1011 kg/m^3
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In seawater
volume of boy = 66/981
so mass of water displacedd submerged =
1029*66/981 = 69.2 kg
BUT
boy only has mass of 68 kg
so he floats
fraction above = (69.2 - 68) /68

a) To determine the density of the boy, we can use the concept of apparent weight in a fluid. The apparent weight is the difference between the weight of an object in air and its weight when submerged in a fluid.

The formula to calculate the apparent weight is given by:

Apparent weight = Weight in air - Buoyant force

The buoyant force can be calculated using the formula:

Buoyant force = Weight of displaced fluid

Since the boy is submerged in water, the weight of the displaced fluid is equal to the weight of the water displaced by the boy's body. The weight of the displaced fluid can be calculated using the formula:

Weight of displaced fluid = Density of fluid * Volume of displaced fluid * gravitational acceleration

In this case, the fluid is water, and the volume of displaced fluid is equal to the volume of the boy's body. The formula to calculate the volume of an object is:

Volume = Mass / Density

We can rearrange the equation to solve for the density:

Density = Mass / Volume

Using the given values, we can now calculate the density of the boy:

Mass of the boy = 68.0 kg
Apparent weight = 6.86 N
Density of water = 1000 kg/m^3
Gravitational acceleration = 9.8 m/s^2

Volume = Mass / Density = 68.0 kg / Density
Buoyant force = Weight of displaced fluid = Density of water * Volume * gravitational acceleration

Apparent weight = Weight in air - Buoyant force
6.86 N = 68.0 kg * 9.8 m/s^2 - Density of water * Volume * 9.8 m/s^2

Substituting the expression for volume and rearranging the equation, we have:

6.86 N = 68.0 kg * 9.8 m/s^2 - Density of water * (68.0 kg / Density) * 9.8 m/s^2

Simplifying the equation:

6.86 N = 667.2 N - 66680 / Density

Rearranging the equation to solve for the density:

66673.14 N = 667.2 N * Density - 6.86 N * Density
66673.14 N = (667.2 N - 6.86 N) * Density
66673.14 N = 660.34 N * Density

Density = 66673.14 N / 660.34 N

Therefore, the density of the boy is:

Density ≈ 101 kg/m^3

b) To determine if the boy will float in sea water, we need to compare the density of the boy to the density of the sea water. If the density of the boy is less than the density of the sea water, he will float. If the density of the boy is greater, he will sink.

Given that the density of sea water is 1029 kg/m^3 and the density of the boy is 101 kg/m^3, the boy's density is less than that of sea water.

Therefore, the boy will float in sea water.

To calculate the percentage of his body volume that will be above sea water, we need to find the ratio of the volume above water to the total volume of the body.

The volume above water can be calculated using the equation:

Volume above water = Total volume of the body - Volume of water displaced

The volume of water displaced is equal to the volume of the body submerged in water, which we calculated earlier using the formula:

Volume = Mass / Density

Since the mass and density of the boy are given, we can calculate the volume submerged in water:

Volume submerged = Mass of the boy / Density of water
Volume submerged = 68.0 kg / 1000 kg/m^3

Knowing the volume submerged, we can now calculate the volume above water:

Volume above water = Total volume of the body - Volume submerged

To calculate the percentage of the body volume above water, we need to divide the volume above water by the total volume of the body and multiply by 100:

Percentage = (Volume above water / Total volume of the body) * 100

Substituting the values, we get:

Volume above water = Total volume of the body - Volume submerged
Percentage = (Volume above water / Total volume of the body) * 100

Calculating the values:

Total volume of the body = Mass of the boy / Density of the boy
Volume above water = (Mass of the boy / Density of the boy) - (Mass of the boy / Density of water)
Percentage = (Volume above water / (Mass of the boy / Density of the boy)) * 100

Substituting the given values:

Total volume of the body = 68.0 kg / 101 kg/m^3
Volume above water = (68.0 kg / 101 kg/m^3) - (68.0 kg / 1000 kg/m^3)
Percentage = (Volume above water / (68.0 kg / 101 kg/m^3)) * 100

Calculating the values:

Total volume of the body ≈ 0.673 m^3
Volume above water ≈ 0.005 m^3
Percentage ≈ (0.005 m^3 / 0.673 m^3) * 100

Therefore, the percentage of the boy's body volume that will be above sea water is approximately:

Percentage ≈ 0.74%

To determine the density of the boy, we need to first understand the concept of density. Density is defined as the mass of an object divided by its volume. In this case, the boy's mass is given as 68.0 kg in air.

a) To find the density of the boy, divide his mass by his volume. The volume can be calculated using the principle of buoyancy. When a body is immersed in a fluid, it experiences an upward force called buoyant force. This buoyant force is equal to the weight of the fluid displaced by the body, which can be calculated using Archimedes' principle:

Buoyant force = weight of fluid displaced by the boy

In this scenario, the apparent weight of the boy underwater is given as 6.86 N. This is the buoyant force acting on the boy, which is equal to the weight of the fluid displaced. Since the boy is completely submerged, the weight of the fluid displaced is equal to his weight in air.

Weight of fluid displaced = weight of the boy = mass of the boy × acceleration due to gravity

Given the mass of the boy as 68.0 kg and acceleration due to gravity as approximately 9.8 m/s², we can calculate the weight of the fluid displaced:

Weight of fluid displaced = 68.0 kg × 9.8 m/s² = 666.4 N

Now we have the weight of the fluid displaced, which is equal to the buoyant force acting on the boy. This buoyant force is also equal to the apparent weight of the boy underwater:

Buoyant force = 6.86 N

Since the density of water is given as 1000 kg/m³, we can use the formula for the density of an object:

Density = mass/volume

We can rearrange the formula for buoyant force to solve for volume:

Buoyant force = Density of water × volume × acceleration due to gravity

Now we can substitute the known values into the equation:

6.86 N = 1000 kg/m³ × volume × 9.8 m/s²

Solving for volume:

Volume = 6.86 N / (1000 kg/m³ × 9.8 m/s²) ≈ 0.0007 m³

Now that we have the volume, we can find the density of the boy by dividing his mass by the volume:

Density = 68.0 kg / 0.0007 m³ ≈ 97,143 kg/m³

Therefore, the density of the boy is approximately 97,143 kg/m³.

b) To determine whether the boy will float in sea water and what percentage of his body volume will be above the sea water, we need to compare the density of the boy with the density of the sea water.

If the density of the boy is less than the density of the sea water, he will float. If the density of the boy is greater than the density of the sea water, he will sink.

Given that the density of sea water is 1029 kg/m³, and the density of the boy is 97,143 kg/m³ (as calculated in part a), we can compare the two densities.

Since the density of the boy is greater than the density of the sea water, he will sink.

The percentage of the boy's body volume above sea water can be calculated using the principle of buoyancy. The volume of the boy that will be above sea water can be found by comparing the difference in densities:

Percentage of boy's body volume above sea water = (Density of sea water / Density of boy) × 100

Substituting the given values:

Percentage of boy's body volume above sea water = (1029 kg/m³ / 97,143 kg/m³) × 100 ≈ 1.06%

Therefore, approximately 1.06% of the boy's body volume will be above sea water.