- 42,793
- 10,493
Looks rightrobax25 said:y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.
The discussion focuses on calculating the motion of a test balloon on a methane planet with a gaseous atmosphere density of 0.1 g/cm³ and a liquid methane density of 0.4 g/cm³. The balloon, with a volume of 1 m³ and a mass of 15 kg, is attached to a 4 kg box by a 10 m rope. The gravitational acceleration on the planet is calculated to be approximately 0.425 m/s², and the buoyant force acting on the box and balloon is derived using Archimedes' principle. The final calculations yield a time of approximately 3.33 seconds for the box to reach the surface of the methane.
PREREQUISITESStudents in physics or engineering, educators teaching fluid dynamics, and anyone interested in gravitational effects and buoyancy in extraterrestrial environments.
Looks rightrobax25 said:y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.
robax25 said:the graph would be like that but I need to change the value.