Balloons are often filled with helium gas because it weighs only about

Tara Alvarado 2022-01-12 Answered
Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as \(\displaystyle{F}_{{{b}}}=\rho_{{{a}{i}{r}}}{g}{V}\) balloon will push the balloon upward. If the balloon has a diameter of 12 m and carries two people, 85 kg each, determine the acceleration of the balloon when it is first released. Assume the density of air is \(\displaystyle\rho={1.16}{k}\frac{{g}}{{m}^{{{3}}}}\), and neglect the weight of the ropes and the cage. Answer: \(\displaystyle{22.4}\frac{{m}}{{s}^{{{2}}}}\)

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Expert Answer

Jonathan Burroughs
Answered 2022-01-13 Author has 2323 answers

Givens and Knowns:
\(\displaystyle\rho_{{{a}}}={1.16}{k}\frac{{g}}{{m}^{{{3}}}}\)
\(\displaystyle{D}={12}{m}\)
\(\displaystyle{m}_{{{p}}}={85}{k}{g}\)
Solution:
\(\displaystyle{a}={\frac{{{F}_{{\ne{t}}}}}{{{m}_{{\to{t}}}}}}\)
\(\displaystyle{F}_{{\ne{t}}}={F}_{{{B}}}-{W}\)
Where \(\displaystyle{F}_{{{B}}}\) is the bouyancy force and W is the total weight.
\(\displaystyle{F}_{{{B}}}=\rho_{{{a}}}\times{V}{o}{l}\times{g}={1.16}\times{\frac{{\pi\times{4}\times{6}^{{{3}}}}}{{{3}}}}\times{9.81}={10296.02}\)
\(\displaystyle{m}_{{\to{t}}}={2}\times{m}_{{{p}}}+\rho_{{{h}{e}}}\times{V}{o}{l}={170}+{\frac{{{1.16}}}{{{7}}}}\times{904.78}={319.934}\)
\(\displaystyle{W}={g}\times{m}_{{\to{t}}}={9.81}\times{319.934}={3138.56}\)
(Note: \(\displaystyle\rho_{{{h}{e}}}={\frac{{\rho_{{{a}}}}}{{{7}}}}\))
Now, we can calculate the net force exerted as follows,
\(\displaystyle{F}_{{\ne{t}}}={10296.02}-{3138.56}={7157.46}\)
\(\therefore a=\frac{7157.46}{319.934}=22.372\)

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einfachmoipf
Answered 2022-01-14 Author has 2973 answers

The mass of the helium is \(\displaystyle{\frac{{{1}}}{{{7}}}}\) (weigth of air)
the radius of the balloon \(\displaystyle{r}={6}{m}\)
then the volume of the balloon is \(\displaystyle{V}={\left({\frac{{{4}}}{{{3}}}}\right)}\pi{r}^{{{3}}}\)
\(\displaystyle={\left({\frac{{{4}}}{{{3}}}}\right)}\pi{\left({6}\right)}^{{{3}}}\)
\(\displaystyle={904.32}{m}^{{{3}}}\)
the mass of each person \(\displaystyle{m}={85}{k}{g}\)
from Newton's law
\(\displaystyle{F}={F}{b}-{F}{g}\)
\(\displaystyle{m}{a}=\text{pair g Vballoon}-{\left[{2}\cdot{85}+{\left({\frac{{{1}}}{{{7}}}}\right)}{\left(\rho{a}{i}{r}\right)}\right]}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}\)
\(\displaystyle={\left({1.16}\right)}{\left({9.8}\right)}{\left({904.32}\right)}-{\left[{2}\cdot{85}+{\left({\frac{{{1}}}{{{7}}}}\right)}{\left({904.32}\right)}{\left({1.16}\right)}\right]}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}\)
\(\displaystyle={10280.3}-{3134.6}={7145.7}\)
\(\displaystyle{\left[{2}\cdot{85}+{\left({\frac{{{1}}}{{{7}}}}\right)}{\left({V}{b}{a}{l}{l}oo{n}\right)}{\left(\rho{a}{i}{r}\right)}\right]}{a}={7145.6}\)
therefore the acceleration is
\(\displaystyle{a}={\frac{{{7145.7}}}{{{319.85}}}}={22.34}\frac{{m}}{{s}^{{{2}}}}\)

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nick1337
Answered 2022-01-14 Author has 10160 answers

\(D=12m \\\rho_{he}=\rho_{air}/7 \rho_{air}=1.16kg/m^{3} \\m_{people}=85 \times 2=170kg \\F_{B}=\rho_{over} Vballon \\\omega=m_{total} \\m_{he}=\rho_{he} Vballon \\V_{balloon}^{2}=\frac{4}{3} \times R_{baloon} \\m_{he}=\frac{1}{7} \times 1.16 \times 904.78=149.93kg \\m_{total}=m_{he}+m_{people}=149.93+170=319.93kg \\\text{Applying Newton's 2nd law.}\ \sum f=m_{total}a \\\therefore F_{B}-\omega=m_{total}a \\\rho_{air} gVballon-m_{total} g=m_{total}a \\a=\frac{\rho_{air}V_{ball}-n_{total}}{m_{total}}g \\=\frac{(1.16)(904.78)-319.93}{319.93} \\=22.4m/s^{2}\)

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