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tensa zangetsu [6.8K]
3 years ago
5

Un’ automobile A di 1,5t percorre un rettilineo a 90km/h. In verso opposto, all’ improvviso, arriva un’ altra automobile B di 0,

88t e la urta frontalmente. Dopo lo scontro il groviglio di rottami continua a spostarsi sulla strada a velocità di 10,2m/s nel verso di A. a) Con quale velocità procedeva l’ automobile B prima dell’ incidente?
Spanish
1 answer:
saul85 [17]3 years ago
7 0

Answer: 15,03 m/s

Explanation:

Dal principio di conservazione della quantità di moto.

m_{1} v_{1}  - m_{2} v_{2}  = (m_{1}  + m_{2} ) v_{3}

Entrambe le auto si stanno muovendo nella direzione opposta, questo si traduce in meno nell'equazione.

Where:

 m_{1}  = Massa dell'auto A in kg = 1,5t = 1500kg

m_{2}  = Massa dell'auto B in kg = 0,88t = 880kg

v_{1} =    Velocità dell'auto A in m / s  = 90km/h = 25 m/s

v_{2} =   Velocità dell'auto B in m / s  

v_{3}  = Velocità comune dopo la collisione in m/s

m_{1} v_{1}  - m_{2} v_{2}  = (m_{1}  + m_{2} ) v_{3}

(1500 x 25) - (880 x v_{2}) = (1500 + 880)10,2

37500 - 880v_{2} = 24276

 -880v_{2} = 24276 - 37500

  -880v_{2} = -13224

    v_{2} = \frac{-13224}{-880}

     v_{2} = 15,03 m/s

 La velocità dell'auto B prima della collisione è di 15,03 m / s

 

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