Answer:
Step-by-step explanation:
In order to solve this problem we first have to use the surface area for the sphere, which is:
[tex]Surface: 4\pi r^{2} \\Surface:4\pi (50)^{2} \\Surface: 31415.93[/tex]
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:
[tex]density=\frac{Mass}{Volume}\\ volume=\frac{Mass}{Density} \\Volume= 37,201.78[/tex]
That is our volume, as we have the area we just divide:
[tex]Volume=Area*Height\\Height=\frac{Volume}{Area}\\ Height=\frac{37201.78}{31415.93} \\Heigth= 1.184 cm[/tex]
So the the height necessary to mantain a neutrally bouyant, with the wieght of the device and the housing would be 1.184 cm