Answer:
∑ τ =0, L₀ = 
Explanation:
In a circular turning movement, when the arms are extended and then contracted in two possibilities:
- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.
In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.
- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted
In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant
∑ τ =0
L₀ =
I₀ w₀ = I w
Answer:
An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.
Explanation:
Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.
So we solve for v:
m*g*h=(1/2)*m*v², masses cancel out,
g*h=(1/2)*v², we multiply by 2,
2*g*h=v² and take the square root to get v
√(2*g*h)=v, we plug in the numbers and get:
v=9.9 m/s.
So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
You are at rest with respect to the car.
You are in motion with respect to the School.
