Question

A block of iron quickly sinks in water, but ships constructed of iron float. A solid cube of iron 1.90 m on each side is made into sheets. From these sheets, to make a hollow cube that will not sink, what should the minimum length of the sides be? (density of iron is 7860 kg/m³)

Answer 1

Answer: 3.78 m

Explanation:

Volume = 1.9 * 1.9 * 1.9

Volume = 6.859 m³

Density of iron assumed to be 7870 kg/m³

Recall,

Density = mass / volume, so

Mass = density * volume

Mass = (7870 * 6.859)

Mass = 53980.33 kg

This mass we calculated, is the mass of water the new cube would have to displace, absolute minimum, so that it can float.

Now density of fresh water is assumed to be (1000 kg/m³).

Thus, the volume of water that will be displaced has to be

Volume = mass / density

Volume = (53980.33 / 1,000)

Volume = 53.98 m³.

Cube root of 53.88 = 3.78 m

Therefore, the minimum side length is 3.78 m

Answer 2

Answer:

3.781488032m or greater

Explanation:

General Rule of thumb is that any thing that is less dense than water will float in it.

So

density of water is 997Kg/m^3 and density of iron on the other hand is  7860kg/m^3.

so to make iron less dense it has to be in larger volume.

mass of the given cube is 53911.74Kg and we have contain it in volume such that density is less than water, mathematically this translates to .

$\frac{53911.74Kg}{V}\leq \frac{997Kg}{m^3}$ left side is density of iron and right side is density of water, solving v in this inequality gives us.

$v \geq 54.0739m^3$ which means each side is $s \geq 3.7814880m$. so the minum is just greater than that.