A 200-L iron tank has a mass of 36 kg. a) Will it float on seawater if empty? b) What mass of potatoes could be added such that the tank would still float? [Note: 1000 L = 1 m3 ]

density = g/mL = 36,000/200,000 = 0.18. It floats.

We can add mass until the density is 1.00 g/mL.
1.0 = mass/volume
V*1.0 = mass = 200,000 1 = 200,000 g
The barge has a mass of 36,000; therefore, 200,000-36,000 = 164,000g = mass to be added. Check to see density = 1.0 g/L. d = m/v = 200,000/200,000 = 1.0 g/mL.

To determine whether the iron tank will float on seawater, we need to compare the density of the tank with the density of seawater.

a) Will the iron tank float on seawater if empty?
The density of an object is calculated by dividing its mass by its volume. In this case, we have the mass of the tank but need to determine its volume.

Since 1000 liters is equal to 1 cubic meter (m³), we can convert the volume of the tank to cubic meters by dividing it by 1000:
Volume of the iron tank = 200 L ÷ 1000 = 0.2 m³

Now we can calculate the density of the iron tank:
Density = Mass / Volume = 36 kg / 0.2 m³ ≈ 180 kg/m³

Seawater has an average density of about 1025 kg/m³.

Comparing the density of the empty tank (180 kg/m³) with the density of seawater (1025 kg/m³), we can see that the tank's density is lower than that of seawater. Therefore, the empty iron tank will float on seawater.

b) What mass of potatoes could be added such that the tank would still float?
To determine the mass of potatoes that could be added to the tank without sinking, we need to find the maximum additional mass that would not exceed the density of seawater (1025 kg/m³).

Let's assume the volume of potatoes that can be added is represented by Vpotatoes, and the density of potatoes is ρpotatoes. The additional mass of potatoes (mpotatoes) can be calculated using the formula:

mpotatoes = Vpotatoes × ρpotatoes

Rearranging the formula, we have:

mpotatoes / Vpotatoes = ρpotatoes

Since we want the density of potatoes to be less than or equal to the density of seawater (1025 kg/m³):

ρpotatoes ≤ 1025 kg/m³

Assuming the density of potatoes is 1000 kg/m³ as a reasonable approximation, we can use this value to calculate the maximum volume of potatoes that can be added:

Vpotatoes ≤ mpotatoes / ρpotatoes = 36 kg / 1000 kg/m³ = 0.036 m³

Converting the volume from cubic meters to liters:
Vpotatoes = 0.036 m³ × 1000 = 36 L

Therefore, a maximum mass of 36 kg of potatoes can be added to the iron tank without causing it to sink.