A shipwrecked mariner is stranded on a desert island. He seals a plea for rescue in a 1.00 liter bottle, corks it up, and throws it into the sea. If the mass of the bottle, plus the message and the air inside, is 0.454 kg, what percentage of the volume of the bottle is submerged as it bobs away? Take the density of seawater to be 1030 kg/m3. For simplicity, assume the bottle and its contents have a uniform density.

Question

A shipwrecked mariner is stranded on a desert island. He seals a plea for rescue in a 1.00 liter bottle, corks it up, and throws it into the sea. If the mass of the bottle, plus the message and the air inside, is 0.454 kg, what percentage of the volume of the bottle is submerged as it bobs away? Take the density of seawater to be 1030 kg/m3. For simplicity, assume the bottle and its contents have a uniform density.

Answer 1

Convert density of bottle and water to

the same units.

Db = 0.454kg/Liter
Db = 0.454kg/1000cm^3
Db = 454.*10^-6kg / cm^3. = Density of bottle.
Vb = 1 Liter = 1000 cm^3. = Vol. of bottle.

Dw = 1030kg/m^3
Dw = 1030kg/10^6cm^3
Dw = 1030*10^-6kg/cm^3. = Density of the water.

Vs = (Db/Dw)* Vb
Vs=(454*10^-6/1030*10^*1000cm^3=441cm^3
Vol. submerged.

%Vs=(Vs/Vb)*100%=(441/1000)*100%=44.1%.

Answer 2

To find the percentage of the volume of the bottle that is submerged, we need to determine the volume of the bottle that is submerged in seawater.

Let's start by finding the buoyant force acting on the bottle. The buoyant force is equal to the weight of the seawater displaced by the submerged portion of the bottle. We can calculate this using Archimedes' principle.

The weight of the seawater displaced is equal to the density of seawater multiplied by the volume of the submerged portion of the bottle.

The density of seawater is given as 1030 kg/m^3, and the volume of the bottle is 1.00 liter, which is equivalent to 0.001 m^3.

So, the weight of the seawater displaced is:
Weight = density * volume
Weight = 1030 kg/m^3 * 0.001 m^3
Weight = 1.03 kg

Since buoyant force equals the weight of the seawater displaced, the buoyant force acting on the bottle is also 1.03 kg.

Now, let's determine the weight of the bottle, message, and air inside. It is given as 0.454 kg.

To calculate the percentage of the volume submerged, we need to find the ratio of the buoyant force to the weight of the bottle, message, and air inside.

Percentage submerged = (Buoyant force / Weight of the bottle, message, and air inside) * 100

Percentage submerged = (1.03 kg / 0.454 kg) * 100
Percentage submerged ≈ 226.76%

Therefore, the percentage of the volume of the bottle submerged as it bobs away is approximately 226.76%.