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A bullet is fired through a board 13.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. The initial speed of the bullet is 560 m/s and it emerges from the other side of the board with a speed of 460 m/s. (a) Find the acceleration of the bullet as it passes through the board.

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Answer:

392307.6923 m/s²

Explanation:

t = Time taken

u = Initial velocity = 560 m/s

v = Final velocity = 460 m/s

s = Displacement = 13 cm

a = Acceleration

From the equation of motion we have

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{460^2-560^2}{2\times 13\times 10^{-2}}\\\Rightarrow a=-392307.6923\ m/s^2[/tex]

The acceleration of the bullet as it passes through the board is -392307.6923 m/s²

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