You are part of a mission to Mars and have been assigned the task of designing balloons for the purpose of carrying scientific instruments over the surface of the planet. You have been told that the density of the atmosphere at the surface is 0.0154 kg m3 . You have with you a tough plastic that has a mass of 5.0 grams per square meter. You plan to use the plastic to construct the balloons and inflate them with hydrogen gas. What would be the radius of a balloon that could hover over the surface

Respuesta :

Answer:

The radius of the balloon is R=  0.974 m

Explanation:

Atmospheric density on the surface of mars ρ  = 0.0154 kg/m³

Mass  density of balloon σ  5 gm/m² = 0.005 kg/m²

Total mass of balloon M  = density of balloon* surface area

The balloon is considered as sphere

Surface area of sphere is given by  4π R²

Mass of balloon is M =σ *4πR²

                      M= (0.005 kg/m²)*(4π R²)

Volume of balloon is given by  [tex]\frac{4}{3}[/tex] πR³

Density of atmosphere of mass is given by  

   ρ = Mass of balloon/volume of  balloon

       =(0.005 kg/m²)*(4π R²)/  [tex]\frac{4}{3}[/tex] πR³

Re arranging R

R =  [tex]\frac{3*0.005 kg/m²}{ 0.0154 kg/m³}[/tex]

   R = 0.974 m

The radius of the balloon is R=  0.974 m

Archimedes' principle allows to find the result for the radius of the balloon that floats in the atmosphere of Mars is:

           R = 0.974 m

   

Given parameters

  • The density of the atmosphere of Mars ρ= 0.0151 kg / m³
  • The mass of the plastic per square meter m = 5.0 g = 5 10⁻³ kg

To find

  • The radius of the globe.

Archimedes' principle states that the thrust on a body is equal to the weight of the dislodged fluid.

           B = ρ g  V

Newton's second law gives a relationship between the force, the mass and the acceleration of the body, in the case that the acceleration is zero, it is called the equilibrium condition.

          Σ F = 0

In the attached we see a diagram of the forces in the system, at the point where the balloon floats the acceleration is zero.

         B -W = 0

         B = W

The surface density is defined by the relation.

            [tex]\sigma = \frac{m}{A}[/tex]

in area is a sphere is:

           A = 4π R²          

           m = 4π R² σ

Indicate that the mass and the area therefore.

          σ = 5 10⁻³ / 1

          σ = 5 10⁻³  kg/m²

The volume of the sphere is:

       V = [tex]\frac{4}{3}[/tex]  π R³

Let's substitute.

       ρ g [tex]\frac{4}{3}[/tex]  π R³ = 4π R² σ g

       [tex]\rho \ \frac{R}{3} = \sigma[/tex]  

        [tex]R = \frac{3 \sigma}{\rho}[/tex]

Let's calculate.

         R = [tex]\frac{3 \ 5 \ 10^{-3} }{0.0154}[/tex]  

         R = 0.974 m

In conclusion using Archimedes' principle we can find the result for the radius of the balloon that floats in the atmosphere of Mars is:

           R = 0.974 m

Learn more here:  brainly.com/question/13106989

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