Respuesta :
[tex]\frac{1}{2}[/tex]Answer:
  [tex]v_{exit } < v_{enter}[/tex]
Explanation:
To propose the solution of this problem we must use the relationship between work and energy
     W = ΔK
the change in kinetic energy is
     ΔK = K_f -K₀
     ΔK = ½ m v_f² - ½ m v₀² = ½ m (v_f² - v₀²)
in this case
     ΔK =  [tex]\frac{1}{2} m ( v_{exit}^2 - v_{enter}^2 )[/tex]
we can find work with the first law of thermodynamics
     [tex]\Delta E_{int}[/tex] = Q + W
where \Delta E_{int} is the internal energy of the body, usually measured in the form of an increase in the temperature of the system
     W = \Delta E_{int} - Q
if we consider that the internal energy does not change
     W = -Q
we substitute everything in the first equation
    -Q =  [tex]\frac{1}{2} m ( v_{exit}^2 - v_{enter}^2 )[/tex]
Because they are squared, the variables are positive, therefore, for the equation to be fulfilled, the exit velocity must be less than the entrance velocity.
      [tex]v_{exit } < v_{enter}[/tex]
