Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>