Your constantly using your body to move around
Answer:
It must be 4 times high.
Explanation:
- Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
- This means, that at any time, the following must be true:
- ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)
⇒ 
- Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.
So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
(1)
- Now, let's do v₂ = 2* v₁
- Replacing in (1) we get:
(2)
- Dividing (2) by (1), and rearranging terms, we get:
- h₂ = 4* h₁
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 
Answer:
The necessary information is if the forces acting on the block are in equilibrium
The coefficient of friction is 0.577
Explanation:
Where the forces acting on the object are in equilibrium, we have;
At constant velocity, the net force acting on the particle = 0
However, the frictional force is then given as
F = mg sinθ
Where:
m = Mass of the block
g = Acceleration due to gravity and
θ = Angle of inclination of the slope
F = 5×9.81×sin 30 = 24.525 N
Therefore, the coefficient of friction is given as
24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479
μ × 42.479 N= 24.525 N
∴ μ = 24.525 N ÷ 42.479 N = 0.577