Respuesta :
Answer:
[tex]n =26786\,ballons[/tex]
Explanation:
According to the Archimedes' Principle, the buoyant force exerted on the helium ballons is equal to the weight of displaced air. A simplified scheme for the system person-ballons is presented below. The correspondent equation of equilibrium is:
[tex]\Sigma F = \rho_{Air}\cdot V_{net}\cdot g - \rho_{He}\cdot V_{net}\cdot g - m_{person}\cdot g = 0[/tex]
[tex](\rho_{Air}-\rho_{He})\cdot V_{net} - m_{person} = 0[/tex]
The minimum net volume is:
[tex]V_{net} = \frac{m_{person}}{\rho_{Air}-\rho_{He}}[/tex]
[tex]V_{net} = \frac{75\,kg}{1.2\,\frac{kg}{m^{3}}-1\,\frac{kg}{m^{3}} }[/tex]
[tex]V_{net} = 375\,m^{3}[/tex]
The volume of each ballon is:
[tex]V_{ballon} = \frac{4}{3}\pi\cdot (0.15\,m)^{3}[/tex]
[tex]V_{ballon} \approx 0.014\,m^{3}[/tex]
The minimum quantity of ballons needed to lift Josiah is:
[tex]n = \frac{V_{net}}{V_{ballon}}[/tex]
[tex]n = \frac{375\,m^{3}}{0.014\,m^{3}}[/tex]
[tex]n =26786\,ballons[/tex]

Answer:
a person with 75 kg mass will need 4762 ballons
Explanation:
The expression for the required number of balloons is
[tex]N = \frac{m}{(\frac{4}{3}\pi r^3) } (p_{air}-p_{he})[/tex]
substitute 1.293kg/m^3 for density of air
0.179kg/m^3 for density of helium
75kg for mass m
radius = diameter / 2
= 0.3m / 2
= 0.15m
[tex]N = \frac{75}{(\frac{4}{3} \pi r^3)(P_{air}-P_{he})}[/tex]
[tex]N = \frac{75}{\frac{4}{3}\pi (0.15)^3(1.293-0.179) }[/tex]
[tex]N = \frac{75}{1.33\pi (3.375\times 10^-^3)(1.114)}[/tex]
[tex]N = \frac{75}{4.178(3.375\times 10^-^3)(1.114)} \\\\N = 4762ballons[/tex]
Hence, a person with 75 kg mass will need 4762 ballons