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vladimir1956 [14]

3 years ago

12

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?

1 answer:

harina [27]3 years ago

6 0

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

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A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

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Answer:

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a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

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$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

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$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

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$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

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$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

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Answer:

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8 0

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