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2 years ago

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A spool of thread has an average radius of 1.00 cm. If the spool contains 62.8 m of thread, how many turns of thread are on the

spool? "Average radius" allows us to not need to treat the layering of threads on lower layers.

1 answer:

Simora [160]2 years ago

7 0

To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:

$N = \frac{l}{\phi}$

Where,

l = length of the thread

$\phi$ = circumference of spool

For \phi we have that,

$\phi = 2\pi r \rightarrow 2\pi (0.01)$

For l we have that

l = 62.8m

Finally the number of Turns would be,

$N = \frac{l}{\phi}$

$N = \frac{62.8}{2\pi (0.01)}$

$N = 1000turns$

Therefore the number of turns of thread on the spool are 1000turns.

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Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
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part (a)

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part (b)

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Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt$

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The lever arm could decrease or increase depending of the initial angle.

Explanation:

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Answer:

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Explanation:

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By differentiating with respect to time t

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Here x is constant that is why

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