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qaws [65]
2 years ago
12

A plane starts from rest accelerates to 40 m/s in 10 seconds. How far did the plane travel during this time?

Physics
1 answer:
tatuchka [14]2 years ago
3 0
<h3><u>Answer:</u></h3>

200 m

<h3><u>Explanation:</u></h3>

<u>We are given:</u>

Initial velocity of the plane (u) = 0 m/s

Final velocity of the plane (v) = 40 m/s

Time interval (t) = 10 seconds

Displacement of the plane (s) = x m

<u>Solving for x:</u>

<u>Acceleration of the plane</u>

v = u + at                                                     [<em>First equation of motion</em>]

40 = 0 + a(10)                                              [<em>replacing known variables</em>]

a = 4 m/s²                                                    [<em>dividing both sides by 10</em>]

<u>Displacement of the Plane:</u>

s = ut + 1/2 (at²)                                            [<em>Second equation of motion</em>]

s = (0)(10) + 1/2(4)(10)²                                  [<em>replacing known variables</em>]

s = 200 m

Hence, the Plane covers a distance of 200 m in the given time interval

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Answer:

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L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

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L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

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ω = L₁/(1/2M + m)R²

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ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

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