Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
ANSWER:
D) centripetal acceleration.
STEP-BY-STEP EXPLANATION:
When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.
Therefore, the answer is centripetal acceleration.
Sorry that you got your answer late but the answer is 0.035m
Answer:
T=4.24 N.m
Explanation:
Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot
F=0.5 kg* 9.8 m/
= 4.9 N
and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus
so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N
and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m
anothe way to do it is,
T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result