Lösung 4.1:8

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Because the wheel's circumference is
Because the wheel's circumference is
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{{Displayed math||<math>2\pi\cdot\text{(radius)} = 2\pi\cdot 0\textrm{.}5\ \text{metres} = \pi\ \text{metres}</math>}}
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{{Abgesetzte Formel||<math>2\pi\cdot\text{(radius)} = 2\pi\cdot 0\textrm{.}5\ \text{metres} = \pi\ \text{metres}</math>}}
the wheel turns <math>\pi</math> metres on each revolution, and in 10 metres the wheel therefore turns
the wheel turns <math>\pi</math> metres on each revolution, and in 10 metres the wheel therefore turns
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{{Displayed math||<math>\frac{10\ \text{metres}}{\pi\ \text{metres}} = \frac{10}{\pi }\ \textrm{revolutions}\approx 3\textrm{.}2\ \textrm{revolutions.}</math>}}
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{{Abgesetzte Formel||<math>\frac{10\ \text{metres}}{\pi\ \text{metres}} = \frac{10}{\pi }\ \textrm{revolutions}\approx 3\textrm{.}2\ \textrm{revolutions.}</math>}}

Version vom 08:49, 22. Okt. 2008

Because the wheel's circumference is

\displaystyle 2\pi\cdot\text{(radius)} = 2\pi\cdot 0\textrm{.}5\ \text{metres} = \pi\ \text{metres}

the wheel turns \displaystyle \pi metres on each revolution, and in 10 metres the wheel therefore turns

\displaystyle \frac{10\ \text{metres}}{\pi\ \text{metres}} = \frac{10}{\pi }\ \textrm{revolutions}\approx 3\textrm{.}2\ \textrm{revolutions.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:8“