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]
