anonymous
  • anonymous
a
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
is the initial speed of the bullet: 3*104=312 m/sec?
Michele_Laino
  • Michele_Laino
oops..312 cm/sec?

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Michele_Laino
  • Michele_Laino
Here, 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
  • Michele_Laino
so, dividing both sides by d, we get: \[\Large F = \frac{{mv_0^2}}{{2d}} = \frac{{0.002 \times {{3.12}^2}}}{{2 \times 0.05}} = ...Newtons\]
Michele_Laino
  • Michele_Laino
that's right!

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