Answer:
So the acceleration of the child will be
Explanation:
We have given angular speed of the child
Radius r = 4.65 m
Angular acceleration
We know that linear velocity is given by
We know that radial acceleration is given by
Tangential acceleration is given by
So total acceleration will be
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
Answer:
Power output: W=1426.9MW
Explanation:
The power output of the falls is given mainly by its change in potential energy:
The potential energy for any point can be calculated as:
If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:
If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:
Answer:
Change in kinetic energy is ( 26CL³)/3
Explanation:
Given :
Net force applied, F(x) = Cx² ....(1)
Displacement of the particle from xi = L to xf = 3L.
The work-energy theorem states that change in kinetic energy of the particle is equal to the net amount of work is done to displace the particle.
That is,
ΔK = W = ∫F·dx
Substitute equation (1) in the above equation.
ΔK = ∫Cx²dx
The limit of integration from xi = L to xf = 3L, so
Substitute the values of xi and xf in the above equation.