Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical

Response:

v = 4.08 m/s²

Clarification:

If my calculations are accurate, the angle is 67.5 degrees.

K=E[(m+M)/M] Kmin=4.4

The total duration from the stone's drop to the echo being heard is 8.9 seconds. The sound takes 0.9 seconds to reach the listener, meaning it takes the stone 8 seconds to descend to the well's bottom. Using gravitational acceleration, the well's depth calculates to be approximately 313.6 meters.

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm