Answer: -1.27 m/s^2
Explanation:
a = - V^2 / 2x
a = -(25^2) / 2 x (246) = 1.27 m/ s^2
Therefore the linear acceleration of the wheel is - 1.27 m/s^2
Answer:
a= (-g) from the moment the ball is thrown, until it stops in the air.
a = (0) when the ball stops in the air.
a = (g) since the ball starts to fall.
Explanation:
The acceleration is <em>(-g)</em> <em>from the moment the ball is thrown, until it stops in the air</em> because the movement goes in the opposite direction to the force of gravity. In the instant <em>when the ball stops in the air the acceleration is </em><em>(0)</em> because it temporarily stops moving. Then, <em>since the ball starts to fall, the acceleration is </em><em>(g)</em><em> </em>because the movement goes in the same direction of the force of gravity
Answer:
The downward force of gravity and the force exerted by the air.
Explanation:
Gravity, in a Newtonian Framework, acts at distance, so, the ball doesn't has to be connected to Earth to feel its gravitational pull (as we know by experience).
The racket only can exert a force in the ball when is touching it, so, once the ball has left the contact with the racket, there is no force by the "hit".
The air is always touching the ball on the atmosphere (even when physicist pretend that is not). So, there is drag from the wind, and a buoyant force exerted by the air over the ball at every time.
