Kraft an seriell geschalteten Federn
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
\(\LaTeX\)
Need help? Yes, please!
The following quantities appear in the problem:
Kraft \(F\) / Strecke \(s\) / Federkonstante \(D\) /
The following formulas must be used to solve the exercise:
\(F = Ds \quad \) \(\frac{1}{D} = \frac{1}{D_1} + \frac{1}{D_2} \quad \)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
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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:
Wie stark müsste man an einer Kombination aus zwei seriell geschalteten Federn mit .millinewtonpercentimeter bzw. .centinewtonpermillimeter Federhärte ziehen damit sich diese um .dm verlängern würde?
Solution:
NewQtyDe.Npm NewQtyDzNpm NewQtyx.m Geg D_ .millinewtonpercentimeter De D_ .centinewtonpermillimeter Dz x .dm x GesZug-KraftFsiN Die Ersatzfederkonstante beträgt: SolQtyD/DeX + /DzX**-Npm D leftfracD_+fracD_right^- fracD_D_D_+D_ leftfracDe+fracDzright^- D Es bräuchte also eine Kraft von: SolQtyFDX * xXN F Dx fracD_D_D_+D_ x fracDe DzDe + Dz x F F fracD_D_D_+D_ x F
Wie stark müsste man an einer Kombination aus zwei seriell geschalteten Federn mit .millinewtonpercentimeter bzw. .centinewtonpermillimeter Federhärte ziehen damit sich diese um .dm verlängern würde?
Solution:
NewQtyDe.Npm NewQtyDzNpm NewQtyx.m Geg D_ .millinewtonpercentimeter De D_ .centinewtonpermillimeter Dz x .dm x GesZug-KraftFsiN Die Ersatzfederkonstante beträgt: SolQtyD/DeX + /DzX**-Npm D leftfracD_+fracD_right^- fracD_D_D_+D_ leftfracDe+fracDzright^- D Es bräuchte also eine Kraft von: SolQtyFDX * xXN F Dx fracD_D_D_+D_ x fracDe DzDe + Dz x F F fracD_D_D_+D_ x F
Meta Information
Exercise:
Wie stark müsste man an einer Kombination aus zwei seriell geschalteten Federn mit .millinewtonpercentimeter bzw. .centinewtonpermillimeter Federhärte ziehen damit sich diese um .dm verlängern würde?
Solution:
NewQtyDe.Npm NewQtyDzNpm NewQtyx.m Geg D_ .millinewtonpercentimeter De D_ .centinewtonpermillimeter Dz x .dm x GesZug-KraftFsiN Die Ersatzfederkonstante beträgt: SolQtyD/DeX + /DzX**-Npm D leftfracD_+fracD_right^- fracD_D_D_+D_ leftfracDe+fracDzright^- D Es bräuchte also eine Kraft von: SolQtyFDX * xXN F Dx fracD_D_D_+D_ x fracDe DzDe + Dz x F F fracD_D_D_+D_ x F
Wie stark müsste man an einer Kombination aus zwei seriell geschalteten Federn mit .millinewtonpercentimeter bzw. .centinewtonpermillimeter Federhärte ziehen damit sich diese um .dm verlängern würde?
Solution:
NewQtyDe.Npm NewQtyDzNpm NewQtyx.m Geg D_ .millinewtonpercentimeter De D_ .centinewtonpermillimeter Dz x .dm x GesZug-KraftFsiN Die Ersatzfederkonstante beträgt: SolQtyD/DeX + /DzX**-Npm D leftfracD_+fracD_right^- fracD_D_D_+D_ leftfracDe+fracDzright^- D Es bräuchte also eine Kraft von: SolQtyFDX * xXN F Dx fracD_D_D_+D_ x fracDe DzDe + Dz x F F fracD_D_D_+D_ x F
Contained in these collections
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Federgesetz und Serieschaltung by TeXercises
| Title | Matched on |
|---|---|
| Kombinierte Federn | tagsformula |
| Seriell geschaltete Federn | tagstitle |
| Verlängerung seriell geschalteter Federn | tagsformula |
| Seriell geschaltete Federn | tagstitle |
| Verlängerung seriell geschalteter Federn | tagsformula |
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