This is the answer to Question 5
Answer:
a = 2 m/s^2
which agrees with the third answer option provided.
Explanation:
Recall the kinematic formula for displacement under the action of a constant acceleration "a":
yf - yi = 1/2 a t^2
using the information provided this equation becomes:
9 = 1/2 a (3)^2
solve for a:
9 * 2 / 9 = a
then a = 2 m/s^2
which agrees with the third answer option provided.
Pretty sure the answer is C. Screw: wheel and axle
Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.
I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
