5 gegen 1
About points...
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".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
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.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
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 -
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 -
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Fünf gleiche Massen M seien wie in der Abbildung gezeigt gleichmässig auf einem Halbkreisbogen mit dem Radius R verteilt. Eine Masse m befinde sich im Zentrum des Kreisbogens. Es sei M kg mkg und Rcm. Wie gross ist die Kraft die die fünf Massen auf m ausüben? center tikzpicturescale. % Koordinatensystem draw thick-latex -- node right x; draw thick-latex -- node above y; % Kreisbogen draw very thick drawblue - arc ::cm; % Radius draw - thick -- node right R .; % Massen M shadedraw shadingballball colorreddrawnone - circle .cm node left M; shadedraw shadingballball colorreddrawnone -.. circle .cm node left M; shadedraw shadingballball colorreddrawnone circle .cm node abovexshiftmm M; shadedraw shadingballball colorreddrawnone .. circle .cm node right M; shadedraw shadingballball colorreddrawnone circle .cm node rightyshiftmm M; % Masse m shadedraw shadingballdrawnone circle .cm node belowyshift-mm m; tikzpicture center
Solution:
Nun aus Symmetriegründen ist die Kraft in x-Richtung: F_resx . Damit ist die resultiere Kraft auf die Masse m: F_res F_resy. Nummerieren wir die Massen von links nach rechts durch dann erhalten wir: F_resy F_sinalpha + F_ + F_sinalpha da F_ F_ F_ F_G und singrad fracsqrt sind gilt: F_resy F_Gsqrt + . Die Kraft ist: F_G GfracMmR^ apx ^-N myRarrow F_res apx .^-N.
Fünf gleiche Massen M seien wie in der Abbildung gezeigt gleichmässig auf einem Halbkreisbogen mit dem Radius R verteilt. Eine Masse m befinde sich im Zentrum des Kreisbogens. Es sei M kg mkg und Rcm. Wie gross ist die Kraft die die fünf Massen auf m ausüben? center tikzpicturescale. % Koordinatensystem draw thick-latex -- node right x; draw thick-latex -- node above y; % Kreisbogen draw very thick drawblue - arc ::cm; % Radius draw - thick -- node right R .; % Massen M shadedraw shadingballball colorreddrawnone - circle .cm node left M; shadedraw shadingballball colorreddrawnone -.. circle .cm node left M; shadedraw shadingballball colorreddrawnone circle .cm node abovexshiftmm M; shadedraw shadingballball colorreddrawnone .. circle .cm node right M; shadedraw shadingballball colorreddrawnone circle .cm node rightyshiftmm M; % Masse m shadedraw shadingballdrawnone circle .cm node belowyshift-mm m; tikzpicture center
Solution:
Nun aus Symmetriegründen ist die Kraft in x-Richtung: F_resx . Damit ist die resultiere Kraft auf die Masse m: F_res F_resy. Nummerieren wir die Massen von links nach rechts durch dann erhalten wir: F_resy F_sinalpha + F_ + F_sinalpha da F_ F_ F_ F_G und singrad fracsqrt sind gilt: F_resy F_Gsqrt + . Die Kraft ist: F_G GfracMmR^ apx ^-N myRarrow F_res apx .^-N.
Meta Information
Exercise:
Fünf gleiche Massen M seien wie in der Abbildung gezeigt gleichmässig auf einem Halbkreisbogen mit dem Radius R verteilt. Eine Masse m befinde sich im Zentrum des Kreisbogens. Es sei M kg mkg und Rcm. Wie gross ist die Kraft die die fünf Massen auf m ausüben? center tikzpicturescale. % Koordinatensystem draw thick-latex -- node right x; draw thick-latex -- node above y; % Kreisbogen draw very thick drawblue - arc ::cm; % Radius draw - thick -- node right R .; % Massen M shadedraw shadingballball colorreddrawnone - circle .cm node left M; shadedraw shadingballball colorreddrawnone -.. circle .cm node left M; shadedraw shadingballball colorreddrawnone circle .cm node abovexshiftmm M; shadedraw shadingballball colorreddrawnone .. circle .cm node right M; shadedraw shadingballball colorreddrawnone circle .cm node rightyshiftmm M; % Masse m shadedraw shadingballdrawnone circle .cm node belowyshift-mm m; tikzpicture center
Solution:
Nun aus Symmetriegründen ist die Kraft in x-Richtung: F_resx . Damit ist die resultiere Kraft auf die Masse m: F_res F_resy. Nummerieren wir die Massen von links nach rechts durch dann erhalten wir: F_resy F_sinalpha + F_ + F_sinalpha da F_ F_ F_ F_G und singrad fracsqrt sind gilt: F_resy F_Gsqrt + . Die Kraft ist: F_G GfracMmR^ apx ^-N myRarrow F_res apx .^-N.
Fünf gleiche Massen M seien wie in der Abbildung gezeigt gleichmässig auf einem Halbkreisbogen mit dem Radius R verteilt. Eine Masse m befinde sich im Zentrum des Kreisbogens. Es sei M kg mkg und Rcm. Wie gross ist die Kraft die die fünf Massen auf m ausüben? center tikzpicturescale. % Koordinatensystem draw thick-latex -- node right x; draw thick-latex -- node above y; % Kreisbogen draw very thick drawblue - arc ::cm; % Radius draw - thick -- node right R .; % Massen M shadedraw shadingballball colorreddrawnone - circle .cm node left M; shadedraw shadingballball colorreddrawnone -.. circle .cm node left M; shadedraw shadingballball colorreddrawnone circle .cm node abovexshiftmm M; shadedraw shadingballball colorreddrawnone .. circle .cm node right M; shadedraw shadingballball colorreddrawnone circle .cm node rightyshiftmm M; % Masse m shadedraw shadingballdrawnone circle .cm node belowyshift-mm m; tikzpicture center
Solution:
Nun aus Symmetriegründen ist die Kraft in x-Richtung: F_resx . Damit ist die resultiere Kraft auf die Masse m: F_res F_resy. Nummerieren wir die Massen von links nach rechts durch dann erhalten wir: F_resy F_sinalpha + F_ + F_sinalpha da F_ F_ F_ F_G und singrad fracsqrt sind gilt: F_resy F_Gsqrt + . Die Kraft ist: F_G GfracMmR^ apx ^-N myRarrow F_res apx .^-N.
Contained in these collections: