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anonymous
 one year ago
a
anonymous
 one year ago
a

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1is the initial speed of the bullet: 3*104=312 m/sec?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops..312 cm/sec?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Here, the work done by the friction force, has to be equal to the kinetic energy of the bullet, so we can write this: \[\Large F \times d = \frac{1}{2}mv_0^2\] where F is the requested force, d= 0.05 meters, and m=0.002 Kg, v_0=3.12 m/sec

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, dividing both sides by d, we get: \[\Large F = \frac{{mv_0^2}}{{2d}} = \frac{{0.002 \times {{3.12}^2}}}{{2 \times 0.05}} = ...Newtons\]
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