Question

A balloon filled with helium gas has an average density of rhob = 0.22 kg/m3. The density of the air is about rhoa = 1.23 kg/m3. The volume of the balloon is Vb = 0.075 m3. The balloon floats upward with acceleration a. 50% Part (a) Express the acceleration of the balloon a in terms of the variables given in the problem statement, assuming upwards is the positive direction. 50% Part (b) Calculate the numerical value of a in m/s2.

Answer 1

Answer:

$a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

45.03681 m/s²

Explanation:

$F_b$ = Buoyant force

W = Weight of the balloon

$\rho_a$ = Density of air = 1.23 kg/m³

$\rho_b$ = Density of balloon = 0.22 kg/m³

$v_a$ = Volume of air

$v_b$ = Volume of balloon

$F_b=\rho_av_bg$

$W=\rho_bv_bg$

g = Acceleration due to gravity = 9.81 m/s²

The net force acting on the balloon is

$F=F_b-W\\\Rightarrow F=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=v_bg(\rho_a-\rho_b)\\\Rightarrow a=\frac{g}{\rho_b}(\rho_a-\rho_b)\\\Rightarrow a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

The equation is $a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

$a=g\left(\frac{\rho_a}{\rho_b}-1\right)\\\Rightarrow a=9.81\times \left(\frac{1.23}{0.22}-1\right)\\\Rightarrow a=45.03681\ m/s^2$

The acceleration of the balloon is 45.03681 m/s²