In a transverse wave, the particles are disturbed in a direction perpendicular to the direction of wave propagation. Thus, waves travel through a medium with no net displacement of the distance between two successive particles of wave that are in wavelength.
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
Answer:
163.33 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 40 Kg
Height (h) = 25 m
Time (t) = 1 min
Power (P) =..?
Next, we shall determine the energy. This can be obtained as follow:
Mass (m) = 40 Kg
Height (h) = 25 m
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E) =?
E = mgh
E = 40 × 9.8 × 255
E = 9800 J
Finally, we shall determine the power. This can be obtained as illustrated below:
Time (t) = 1 min = 60 s
Energy (E) = 9800 J
Power (P) =?
P = E/t
P = 9800 / 60
P = 163.33 Watts
Thus, the power required is 163.33 Watts
