Problem: A lighter-than-air spherical balloon and its load ofpassengers and ball

Question

Problem: A lighter-than-air spherical balloon and its load ofpassengers and ballast are floating stationary above the earth.Ballast is weight (of negligible volume) that can be droppedoverboard to make the balloon rise. The radius of this balloonis 6.25 m. Assuming a constant value of 1.29 kg/m3for the density of air, determine how much weight must be droppedoverboard to make the balloon rise 105 m in 15.0 s. Since, it's in equilibrium I know the weight of the balloon isequal to the buoyant force = weight of displaced air =1.29*9.81*4/3*PI*6.25^3=12941.6 N Using Y=1/2 a t^2, I can get a = 2*y/t^2 = 2*105/15^2 = 0.9333m/s^2. But, I don't know what to do next. Seems like there issomething missing. A lighter-than-air spherical balloon and its load ofpassengers and ballast are floating stationary above the earth.Ballast is weight (of negligible volume) that can be droppedoverboard to make the balloon rise. The radius of this balloonis 6.25 m. Assuming a constant value of 1.29 kg/m3for the density of air, determine how much weight must be droppedoverboard to make the balloon rise 105 m in 15.0 s. Since, it's in equilibrium I know the weight of the balloon isequal to the buoyant force = weight of displaced air =1.29*9.81*4/3*PI*6.25^3=12941.6 N Using Y=1/2 a t^2, I can get a = 2*y/t^2 = 2*105/15^2 = 0.9333m/s^2. But, I don't know what to do next. Seems like there issomething missing.

Explanation / Answer

You have figured out that acceleration = a = 0.9333ms-2 And upward force = 12941.6 N Now, Let mass of ballast be m Since accn reqd is a... So, ma = upward force. =>m =12914/0.9333          =13836.92275Kg

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.