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9966 [12]
3 years ago
7

A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a s

peed of +370 m/s. The bullet then emerges on the other side of the board with a speed of +259 m/s. Assuming constant acceleration (rather, deceleration!) of the bullet while inside the wooden board, calculate the acceleration.
Remember that since the bullet is traveling in the positive direction and it is slowing down, the acceleration is in the opposite, or negative, direction. (In this case the acceleration is negative.)
Calculate also the total time the bullet is in contact with the board (in sec).
Physics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

-387883.3 m/s²

0.000286168546055 seconds

Explanation:

t = Time taken

u = Initial velocity = 370 m/s

v = Final velocity = 259 m/s

s = Displacement = 9 cm

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2

The acceleration of the bullet is -387883.3 m/s²

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s

The time taken is 0.000286168546055 seconds

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how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at
DiKsa [7]

Answer:

|a|=2.83\ m/s^2

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Explanation:

<u>Net Force And Acceleration </u>

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The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

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