Answer:
the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
Explanation:
This exercise asks to describe the inflation situation of a spherical fultball.
Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.
When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.
Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
I would say a. Good luck!
Answer:
12.25 meters
Explanation:
s=1/2(v+u)t
s= displacement
v= final velocity
u= initial velocity
t= time
7m/s+0m/s divide by 2= 3.5 m/s velocity Times 3.5 seconds= 12.25 meters
Answer:
Explanation:
Given that
Force constant k=8.6N/m
Weight =64g=64/1000=0.064kg
Extension is 45mm=45/1000= 0.045m
It will have it highest spend when the Potential energy is zero
Therefore energy in spring =change in kinetic energy
Ux=∆K.e
½ke² = ½mVf² — ½mVi²
Initial velocity is 0, Vi=0m/s
½ke² = ½mVf²
½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²
0.0087075 = 0.032 Vf²
Then, Vf² = 0.0087075/0.032
Vf² = 0.2721
Vf=√0.2721
Vf= 0.522m/s
The time it will have this maximum velocity?
Using equation of motion
Vf= Vi + gr
0.522= 0+9.81t
t=0.522/9.81
t= 0.0532sec
t= 53.2 milliseconds
Answer:
The catcher does negative work on the ball because the force exerted by the catcher is opposite in direction to the motion of the ball.
Explanation:
