Question

A balloon is released from a tall building. The total mass of the balloon including the enclosed gas is 2.0 kg. Its volume is 5.0 m3. The density of air is 1.3 kg/m3.

Will the balloon rise, fall, or remain stationary; and why?




The balloon will fall because the downward buoyant force is greater than the upward buoyant force.



The balloon will fall because the upward buoyant force is less than its weight.



The balloon will fall because its density is greater than that of air.



The balloon will remain stationary because its density is less than that of air.



The balloon will rise because the upward buoyant force is greater than its weight.

Answer 1

Answer on Question#37651, Physics, Other

Question:

A balloon is released from a tall building. The total mass of the balloon including the enclosed gas is $2.0\mathrm{kg}$ . Its volume is $5.0\mathrm{m}^3$ . The density of air is $1.3\mathrm{kg/m}^3$ . Will the balloon rise, fall, or remain stationary; and why?

Answer:

where $F_{b}$ is upward buoyant force, $P$ - weight.

buoyant force equals (assuming Archimedes' principle):

$$F _ {b} = \rho_ {a} g V = 1. 3 \frac {k g}{m ^ {3}} 9. 8 \frac {m}{s ^ {2}} 5 m ^ {3} = 6 3. 7 N$$

where $\rho_{a}$ is density of air, $V$ is volume of the balloon

weight of the balloon equals:

$$P = m g = 2 k g * 9. 8 \frac {m}{s ^ {2}} = 1 9. 6 N$$

where $m$ is total mass of the balloon.

So, we have $F_{b} > P$ , therefore the balloon rise.

Answer: the balloon rise.