A hydrogen balloon, with a volume of 5000m3 and a weight of 15000n, is known to have a density of 0.09kg/m3 for hydrogen and 1.29kg/m3 for air

A hydrogen balloon, with a volume of 5000m3 and a weight of 15000n, is known to have a density of 0.09kg/m3 for hydrogen and 1.29kg/m3 for air

Buoyancy = PGV = 5000 * 1.29 * 10 = 64500n
Self weight + hydrogen weight = 15000 + P hydrogen GV = 15000 + 5000 * 0.09 * 10 = 19500n
So the maximum load is 64500-19500 = 45000n

In a certain period of time, the hydrogen balloon just rises at a constant speed. In this period, its volume is 200 COM3, and the known air density is 1.29 kg / m3
During this period, the buoyancy in the air is () n, and the self weight of the hydrogen balloon is () n, taking g = 10N / kg
As you just analyzed
A solid object is hung under the spring dynamometer. The indication of the spring dynamometer is F. if the object is immersed in water and the indication of the spring dynamometer is five times F when the object is stationary, the density of the object is ()
A.1.0×10^3 kg/m3 B 0.8×10^3 kg/m3
C.1.5×10^3 kg/m3 D.1.25×10^3 kg/m3

200com3=0.0200m3
F = PGV = 1.29kg/m3 * 10N / kg * 0.02m3 = 0.258n
Gravity g = f floating = 0.258n

There is a 0.1m3 ice floating on the water surface (ice = 0.9 × 103kg / m3, water = 0.9 × 103kg / m3, g = 10N)/
When there is a 0.1m3 ice floating on the water surface (ice = 0.9 × 103kg / m3, water = 0.9 × 103kg / m3, g = 10N / kg), the size and process of the ice buoyancy will change