a En los primeros 6 segundos recorre una distancia d 1 dada por d 1 1 2 a t 2 1

A en los primeros 6 segundos recorre una distancia d

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a) En los primeros 6 segundos recorre una distancia d 1 , dada por d 1 = 1 2 a t 2 1 = 1 2 a (6) 2 = 18 a y alcanza su velocidad m´axima v m , dada por v m = a t 1 = 6 a en los restantes 9 segundos recorre una distancia d 2 = v m t 2 = (6 a ) × 9 = 54 a juntando ambas ecuaciones y usando que: 100 = d 1 + d 2 , obtenemos 100 = 18 a + 54 a a = 1 , 39 m / s 2 b) Velocidad m´axima. Ten´ ıamos que v m = 6 a v m = 8 , 33 (m / s) c) Distancia recorrida a los 10 segundos. En los primeros 6 segundos recorre d 1 = 1 2 a t 2 1 = 25 m En los restantes 4 segundos va a su velocidad m´axima y recorre d 3 = 8 , 33 × 4 m = 33 , 33 m de modo que a los 10 segundos ha recorrido d = d 1 + d 3 = 58 , 33 m Pregunta 2. M´aquina de Atwood. En ausencia de roce y si la cuerda tiene masa despreciable, entonces tenemos que para m 1 : m 1 g - T 1 = m 1 a 1 para m 2 : m 2 g - T 2 = m 2 a 2 1
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Donde se ha tomado la direcci´on positiva hacia abajo. Si la polea tiene masa despreciable T 1 = T 2 Si la cuerda es inextensible a 2 = - a 1 , es decir una masa baja y la otra sube. Sumando ambas ecuaciones obtenemos, a 1 = ( m 1 - m 2 ) g ( m 1 + m 2 ) = 4 , 2 m / s 2 b) Velocidad de m 1 luego de haber descendido 0,75 m. Sabemos que v 1 = p 2 a y = 2 , 51 m / s c) Tiempo que demora en descender esos 0,75 m Sabemos que y = 1 2 a t 2 t = 0 , 598 s Pregunta 3.
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  • Summer '20
  • Aceleración, Kilogramo, Cantidad de movimiento, 100 metros, 0,75 m

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