Question

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).

Answer 1

Answer:

The mass of the cargo is $M = 188.43 \ kg$

Explanation:

From the question we are told that

    The radius of the spherical balloon is  $r = 7.40 \ m$

     The mass of the balloon is  $m = 990\ kg$  

The volume of the spherical balloon is mathematically represented as

     $V = \frac{4}{3} * \pi r^3$

substituting values

      $V = \frac{4}{3} * 3.142 *(7.40)^3$

      $V = 1697.6 \ m^3$

The total mass  the balloon can lift is mathematically represented as

     $m = V (\rho_h - \rho_a)$

where $\rho_h$ is the density of helium with a  value of

       $\rho_h = 0.179 \ kg /m^3$

and  $\rho_a$ is the density of air with a value of

        $\rho_ a = 1.29 \ kg / m^3$

substituting values

          $m = 1697.6 ( 1.29 - 0.179)$

         $m = 1886.0 \ kg$

Now the mass of the cargo is mathematically evaluated as

        $M = 1886.0 - 1697.6$

        $M = 188.43 \ kg$