A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 990 kg ? Neglect the buoyant force on the cargo volume itself. Assume gases are at 0∘C and 1 atm pressure (rhoair = 1.29 kg/m3, rhohelium = 0.179 kg/m3).

Respuesta :

Answer:

The mass of the cargo is [tex]M = 188.43 \ kg[/tex]

Explanation:

From the question we are told that

    The radius of the spherical balloon is  [tex]r = 7.40 \ m[/tex]

     The mass of the balloon is  [tex]m = 990\ kg[/tex]  

The volume of the spherical balloon is mathematically represented as

     [tex]V = \frac{4}{3} * \pi r^3[/tex]

substituting values

      [tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]

      [tex]V = 1697.6 \ m^3[/tex]

The total mass  the balloon can lift is mathematically represented as

     [tex]m = V (\rho_h - \rho_a)[/tex]

where [tex]\rho_h[/tex] is the density of helium with a  value of

       [tex]\rho_h = 0.179 \ kg /m^3[/tex]

and  [tex]\rho_a[/tex] is the density of air with a value of

        [tex]\rho_ a = 1.29 \ kg / m^3[/tex]

substituting values

          [tex]m = 1697.6 ( 1.29 - 0.179)[/tex]

         [tex]m = 1886.0 \ kg[/tex]

Now the mass of the cargo is mathematically evaluated as

        [tex]M = 1886.0 - 1697.6[/tex]

        [tex]M = 188.43 \ kg[/tex]