Question

11. Challenge: A certain device must be created to house a scientific instrument. The housing must be a spherical shell, with an outside diameter of 1 m. It will be made of a material whose density is 14 g/cm3. It will house a sensor inside that weighs 1.2 kg. The housing, with the sensor inside, must be neutrally buoyant, meaning that its density must be the same as water. Ignoring any air inside the housing, and assuming that water has a density of 1 g/cm3, how thick should the housing be made so that the device is neutrally buoyant? Round your answer to the nearest tenth of a centimeter.

Answer 1

Answer:

Step-by-step explanation:

In order to solve this problem we first have to use the surface area for the sphere, which is:

$Surface: 4\pi r^{2} \\Surface:4\pi (50)^{2} \\Surface: 31415.93$

We have that the area of the metal would be 31415.93 cm2.

So we need the sphere to have a minor density than 1g/cm3 in order to keep its bouyancy on water, and the volume of the sphere is: 522,025 cm3.

That in kilograms/m3 would be 522.025 kg, we withdraw form this the weight of the device 1.2 kg=1200 g= 520 825 g

Now we just divide the weight by the density to see the volume of our housing:

$density=\frac{Mass}{Volume}\\ volume=\frac{Mass}{Density} \\Volume= 37,201.78$

That is our volume, as we have the area we just divide:

$Volume=Area*Height\\Height=\frac{Volume}{Area}\\ Height=\frac{37201.78}{31415.93} \\Heigth= 1.184 cm$

So the the height necessary to mantain a neutrally bouyant, with the wieght of the device and the housing would be 1.184 cm