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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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Answer:

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The distance travel by the rock is approximately 132.496 m

Part B

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Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

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∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

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Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

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The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
2 years ago
A basketball player passes a ball to a teammate at a velocity of 6 m/s. The ball has a mass of 0.51 kg. If the original player h
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We can solve the problem by using conservation of momentum.

The player + ball system is an isolated system (there is no net force on it), therefore the total momentum must be conserved. Assuming the player is initially at rest with the ball, the total initial momentum is zero:

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The total final momentum is:

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where p_p is the momentum of the player and p_b is the momentum of the ball.

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