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Durchmesser von Metallkugel

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We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
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That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.

About difficulty...

We associate a certain difficulty with each exercise.
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/durchmesser-von-metallkugel/

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Solution

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The following quantities appear in the problem: Masse \(m\) / Volumen \(V\) / Radius \(r\) / Dichte \(\varrho\) /

The following formulas must be used to solve the exercise: \(\varrho = \dfrac{m}{V} \quad \) \(V = \dfrac{4}{3}\pi r^3 \quad \)

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Exercise:
Eine Kugel aus Metall r hat m Masse. Wie gross ist ihr Durchmesser?

Solution:
Das Volumen einer Kugel ist mathematisch gegeben durch: V fracpi r^ Das Volumen eines Körpers können wir physikalisch zudem durch Dichte und Masse ausdrücken: V fracmrho Die beiden Volumen sind natürlich identisch; wir können also die Ausdrücke gleichsetzen und algebraisch nach dem Radius r auflösen womit dann auch der gesuchte Durchmesser d feststeht: fracpi r^ fracmrho r^ fracmrho frac fracpi r^ fracmpirho r leftfracmpirhoright^frac sqrtfracmpirho leftfrac pqkgpi pqkgpmkright^frac pq.m Der Durchmesser ist nun evidenterweise der doppelte Radius drpq.mpq.cm.

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Tags	dichte, durchmesser, kugel, masse, mechanik, metall, physik, volumen
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Difficulty	(3, default)
Points	3 (default)
Language	(Deutsch)
Type	Calculative / Quantity
Creator	uz
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