a hot - air balloon plus cargo has a mass of 1890 kg and a volume of 11,490 m³. the balloon is floating at a…

a hot - air balloon plus cargo has a mass of 1890 kg and a volume of 11,490 m³. the balloon is floating at a constant height of 6.25 m above the ground. what is the density of the hot air in the balloon?\n\na. 7.24 kg/m³\n\nb. 4.83 kg/m³\n\nc. 5.91 kg/m³\n\nd. 1.12 kg/m³\n\ne. 3.22 kg/m³

a hot - air balloon plus cargo has a mass of 1890 kg and a volume of 11,490 m³. the balloon is floating at a constant height of 6.25 m above the ground. what is the density of the hot air in the balloon?\n\na. 7.24 kg/m³\n\nb. 4.83 kg/m³\n\nc. 5.91 kg/m³\n\nd. 1.12 kg/m³\n\ne. 3.22 kg/m³

Answer

Explanation:

Step1: Recall the buoyancy principle

The buoyant force $F_b$ equals the weight of the displaced air, and since the balloon is floating at a constant height, $F_b = mg$. The buoyant force formula is $F_b=\rho_{air}Vg$, where $\rho_{air}$ is the density of the surrounding air (assume standard - density of air $\rho_{air} = 1.29\ kg/m^3$), $V$ is the volume of the balloon, and $g$ is the acceleration due to gravity ($g = 9.8\ m/s^2$). Also, $mg=\rho_{hot - air}Vg+ m_{cargo}g$.

Step2: Rearrange the equation for density of hot - air

We can rewrite $mg=\rho_{hot - air}Vg+ m_{cargo}g$ as $\rho_{hot - air}=\frac{m - m_{cargo}}{V}$. Given $m = 1890\ kg$ and $V = 11490\ m^3$. [ \begin{align*} \rho_{hot - air}&=\frac{1890}{11490}\ &\approx1.12\ kg/m^3 \end{align*} ]

Answer:

d. $1.12\ kg/m^3$