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Masse eines Achterbahnzuges

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That being said... How many "default points" should you associate with an exercise upon creation?
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Again, very vague... But the number should kind of represent the "work" required.

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When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.

That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:

Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise

Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise

Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise

Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise

Level 5 -
Level 6 -

Exercise

https://texercises.com/exercise/masse-eines-achterbahnzuges/

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Solution

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The following quantities appear in the problem: Masse \(m\) / Kraft \(F\) / Energie \(E\) / Geschwindigkeit \(v\) / Ortsfaktor \(g\) / Höhe \(h\) / Radius \(r\) /

The following formulas must be used to solve the exercise: \(F = mg \quad \) \(E_{\rm \scriptscriptstyle kin} = \dfrac12 mv^2 \quad \) \(E_{\rm \scriptscriptstyle pot} = mgh \quad \) \(F = m\dfrac{v^2}{r} \quad \)

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Exercise:
Ein vollbesetzter Wagen eines Achterbahnzuges durchfahre den höchsten Punkt eines Loopings mit hO Höhe mit vO wobei die Kraft der Schienen auf den Wagen FO betrage. Welche Masse errechnet man aus diesen Daten für den Zug?

Solution:
Die Kraft der Schiene und die Gravitation halten die Achterbahn im höchsten Punkt auf der Kreisbahn: sscFS + sscFG msscaZ sscFS + mg mfracv^r sscFS + mg fracmv^frach Aufgelöst nach der Masse findet man: SolQtymFX/*vX^/hX-ncgXkg fracmv^h-mg sscFS m leftfracv^h-gright sscFS m fracsscFSfracv^h-g fracFfracqtyv^h -g m

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Todesspirale by TeXercises

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Asked Quantity: Masse \(m\) in Kilogramm \(\rm kg\)

Attributes & Decorations

Tags	achterbahn, dynamik, geschwindigkeit, kraft, kreisbewegung, zentrifugalkraft, zentripetalkraft
Content image
Difficulty	(4, default)
Points	3 (default)
Language	(Deutsch)
Type	Calculative / Quantity
Creator	uz
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