Question

A balloon used for a physics experiment has a radius of 3.15 m and is filled with helium. The total mass of the balloon is 15.2 kg and the density of the surrounding air is 1.28 kg/m³.
a. How much buoyant force is acting on the balloon?
b. When the ballonon is released from the ground what is the acceleration of the balloon?
c. As the balloon rises what happens to the radius of the balloon? Explain.

Include drawing to explain.

Answer 1

Solutions given m= 15.2kg , k 23,15 m Sa = 1.28 kg/ V 4 R = 4X3.14 X 3,15 X3, 15 X 3.15 V=130.858 m3 chore & is the valom of sphere formoler of booyant force F = Vlag R = 130.858*1.28X9,8 T = 1641.48N epewart direction, (a) = 1641.48 N opwart direction Ans Let acceleration of ballon is do Framg=ma. Ao tong = 1641. 42 - 9,8 Ao=107.992-9. 2 Ao=98. 192 m/s2 (6) Acceleration of ballen is a. = 98. 192 m/s2 upward direction Ans 15.2

(③ Let radios of ballon is R V = 4 R² radius will be effect on the buoyant force F = Vsag where Fo is the beayant force V is the valom of sphere Sa is the density of air Fo= 48 R² Sag upeward acceleration to an F-ng-mao

13m A. = Eng Ao = (4 * L R-1) Effect of the radius on the acceleration. A. = (4th & 1 ) ) effect of the radies on the buoyant Force / 3m Fo= HAR Sag