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FromTheMoon [43]
2 years ago
11

A motorbike is traveling to the left with a speed of 27.0\,\dfrac{\text m}{\text s}27.0 s m ​ 27, point, 0, start fraction, star

t text, m, end text, divided by, start text, s, end text, end fraction when the rider slams on the brakes. The bike skids 41.5\,\text m41.5m41, point, 5, start text, m, end text with constant acceleration before it comes to a stop. What was the acceleration of the motorbike as it came to a stop?
Physics
1 answer:
myrzilka [38]2 years ago
7 0

Answer:

\frac{729\frac{m^{2} }{s^{2} } }{83m}≈8.8\frac{m}{s^{2} }

Explanation:

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A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
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The speed with which the diver is moving, vₓ = 8.0 m/s

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1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

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Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

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v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

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Answer:

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