Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
C. A mechanical wave generally travels faster in gases than liquids.
Answer:
find the sum of the inital and final velocitys and divide by 2 to find the average
The correct answer is C) towards the center of the circle.
Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.
For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.
Answer: Option B. R = (1/2)gt^2
Explanation:
S = R (horizontal distance)
V^2 = 2gS
V^2 = 2gR
R = V^2 / 2g
But V = gt
R = (gt)^2 / 2g
R = (g^2 x t^2) / 2g
R = gt^2 / 2
But t^2 = 2h/g
R = ( g x 2h/g) / 2
R = h
But h = (1/2)gt^2
R = h = (1/2)gt^2
