Answer:
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](https://img.qammunity.org/2020/formulas/mathematics/high-school/6wwpovgad92crp172lx7g3zghtfvxs3zab.png)
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=(Mass)/(Volume)\\ volume=(Mass)/(Density) \\Volume= 37,201.78](https://img.qammunity.org/2020/formulas/mathematics/high-school/52xlmmxqmb1tcwb6w87c6nikmv322cikoj.png)
That is our volume, as we have the area we just divide:
![Volume=Area*Height\\Height=(Volume)/(Area)\\ Height=(37201.78)/(31415.93) \\Heigth= 1.184 cm](https://img.qammunity.org/2020/formulas/mathematics/high-school/ob972akt1918d29xfi1wcpbjnf9zaw9kif.png)
So the the height necessary to mantain a neutrally bouyant, with the wieght of the device and the housing would be 1.184 cm