(2) According to (1), when the hydrogen balloon with radius r moves at speed v, the resistance f is directly proportional to the speed v, so if the volume of the hydrogen balloon remains the same during its ascent, the greater the speed, the greater the resistance.
(3) When the hydrogen balloon rises in the air, it is affected by gravity, buoyancy and resistance.
Gravity on the hydrogen balloon: g = mg = ρ VG = 43 π R3 ρ 0g;
Buoyancy of hydrogen balloon: f float = ρ VG = 43 π R3 ρ g;
According to (1), when the speed of the hydrogen balloon is vT, the resistance of the hydrogen balloon is f=kvTr2.
According to Figure B, when the speed is vT, the hydrogen balloon will rise at a uniform speed, and the resultant force on the hydrogen balloon is zero, so G+f=F floats.
43πr3ρ0g+kvTr2=43πr3ρg
The solution is vt = 4 π r (p? ρ0)g3k
So the answer is: (1); (2) large; (3)4πr(ρ? ρ0)g3k。