How much steel is needed to suspend a balloon at 50 ft below sea level?

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To calculate the weight of steel needed to suspend a 12-inch diameter balloon filled with air at 50 feet below sea level in seawater, understanding buoyancy is essential. The buoyant force can be determined using the density of seawater, which is approximately 1025 kg/m³. The weight of the balloon must be countered by the weight of the steel, which has a density of 7850 kg/m³. The discussion encourages practical application of physics principles, particularly focusing on the buoyancy equation. Estimating the required steel weight involves applying these densities and understanding the forces at play in the underwater environment.
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I need some help with a calculation. I am trying to calculate how much steel in weight would be needed to suspend a ball(essentially a balloon) 12in in diameter full of air at 50 ft below sea level, in sea water. I am not the best at physics and I think I have an idea but wanted a good opinion. Does not have to be exact just a good estimation.
 
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Welcome to PF!

The best way to learn is to learn by doing. So take a shot. Some hints to get you started: do you know the equation for buoyancy? Density of water and steel?
 
seawater=1025 kgm3 Steel = 7850 kgm3 and I am not exactly sure which formula to use?
 

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