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Dmitrij [34]
3 years ago
11

Una pelota de béisbol de 142g de masa, luego de ser arrojada por el pitcher lleva una velocidad de 90mph. Luego de ser bateada s

e mueve en sentido contrario a 54 m/s.
a. Calcular el impulso del bate sobre la pelota.
b. Si la pelota permanece en contacto con el bate 0,008 s ¿cuál es el módulo de
la fuerza del golpe?
Physics
1 answer:
V125BC [204]3 years ago
5 0

Answer:

a)    I = 13.38 kg m / s, b)    F = 1,373 10³ N

Explanation:

The impulse is given by the relation

          I = ∫ F dt = Δp

          I = p_f -p₀

          I = m (v_f - v₀)

take the ball's exit direction as positive, whereby the ball velocities

v₀ = -90mph, the final velocity v_f = + 54 m / s

Let's reduce the units to

         I = 0.142 [54- (-40.23) ]

      the SI system

        v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s

        m = 142 g (1kg / 1000) = 0.142 kg

we calculate  

          I = 0.142 [54- (-40) ]

          I = 13.38 kg m / s

b) let's use the definition of momentum

         I = ∫ F .dt

         I = F ∫ dt

         F = I / t

         F = 13.38 / 0.008

         F = 1,373 10³ N

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