Aluminiumdraht Wärme zuführen
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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.
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
\(\LaTeX\)
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The following quantities appear in the problem:
Zeit \(t\) / Länge \(\ell\) / Masse \(m\) / Temperatur \(T\) / Arbeit \(W\) / Energie \(E\) / Leistung \(P, \Phi\) / Volumen \(V\) / Wärme \(Q\) / Dichte \(\varrho\) / Längenausdehnungskoeffizient \(\alpha\) / Wärmekapazität \(c\) /
The following formulas must be used to solve the exercise:
\(\varrho = \dfrac{m}{V} \quad \) \(Q = c \cdot m \cdot \Delta\vartheta \quad \) \(P = \dfrac{E}{t} = \dfrac{W}{t} = \dfrac{Q}{t} \quad \) \(\Delta \ell = \ell_0 \cdot \alpha \cdot \Delta\theta \quad \) \(V = V_0 \cdot (1+ 3\alpha \cdot \Delta\vartheta) \quad \)
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Exercise:
Einem lO langen Aluminiumdraht rO cO alpO wird währ tO verlustfrei eine Wärmeleistung von PO zugeführt. Um wie viel verlängert sich der Draht falls sein Querschnitt anfänglich AO betrug?
Solution:
Geg ell_ lO l rho rO r c cO c alpha alpO alp t tO t P PO P A AO A GesLängenveränderungDeltaellsim Der Aluminiumdraht eigentlich ein langer/hoher Zylinder hat anfänglich SolQtyVAX*lXcubicmeter V Aell_ A l V Volumen und damit SolQtymrX*VXkg m rho V rho Aell_ r A l m Masse. Führt man ihm SolQtyQPX*tXJ E Q Pt P t Q Wärmeenergie zu wird seine Temperatur um SolQtydTQX/cX/mXcelsius Deltatheta fracQcm fracPtc rho Aell_ fracP tc r A l dT ansteigen womit er sich SolQtydllX*alpX*dTXm Deltaell ell_ alpha Deltatheta ell_ alpha fracPtc rho Aell_ fracalpha t Prho c A fracalp t Pr c A dl approx dlS dlP- verlängern wird. Deltaell fracalpha t Prho c A dlP-
Einem lO langen Aluminiumdraht rO cO alpO wird währ tO verlustfrei eine Wärmeleistung von PO zugeführt. Um wie viel verlängert sich der Draht falls sein Querschnitt anfänglich AO betrug?
Solution:
Geg ell_ lO l rho rO r c cO c alpha alpO alp t tO t P PO P A AO A GesLängenveränderungDeltaellsim Der Aluminiumdraht eigentlich ein langer/hoher Zylinder hat anfänglich SolQtyVAX*lXcubicmeter V Aell_ A l V Volumen und damit SolQtymrX*VXkg m rho V rho Aell_ r A l m Masse. Führt man ihm SolQtyQPX*tXJ E Q Pt P t Q Wärmeenergie zu wird seine Temperatur um SolQtydTQX/cX/mXcelsius Deltatheta fracQcm fracPtc rho Aell_ fracP tc r A l dT ansteigen womit er sich SolQtydllX*alpX*dTXm Deltaell ell_ alpha Deltatheta ell_ alpha fracPtc rho Aell_ fracalpha t Prho c A fracalp t Pr c A dl approx dlS dlP- verlängern wird. Deltaell fracalpha t Prho c A dlP-
Meta Information
Exercise:
Einem lO langen Aluminiumdraht rO cO alpO wird währ tO verlustfrei eine Wärmeleistung von PO zugeführt. Um wie viel verlängert sich der Draht falls sein Querschnitt anfänglich AO betrug?
Solution:
Geg ell_ lO l rho rO r c cO c alpha alpO alp t tO t P PO P A AO A GesLängenveränderungDeltaellsim Der Aluminiumdraht eigentlich ein langer/hoher Zylinder hat anfänglich SolQtyVAX*lXcubicmeter V Aell_ A l V Volumen und damit SolQtymrX*VXkg m rho V rho Aell_ r A l m Masse. Führt man ihm SolQtyQPX*tXJ E Q Pt P t Q Wärmeenergie zu wird seine Temperatur um SolQtydTQX/cX/mXcelsius Deltatheta fracQcm fracPtc rho Aell_ fracP tc r A l dT ansteigen womit er sich SolQtydllX*alpX*dTXm Deltaell ell_ alpha Deltatheta ell_ alpha fracPtc rho Aell_ fracalpha t Prho c A fracalp t Pr c A dl approx dlS dlP- verlängern wird. Deltaell fracalpha t Prho c A dlP-
Einem lO langen Aluminiumdraht rO cO alpO wird währ tO verlustfrei eine Wärmeleistung von PO zugeführt. Um wie viel verlängert sich der Draht falls sein Querschnitt anfänglich AO betrug?
Solution:
Geg ell_ lO l rho rO r c cO c alpha alpO alp t tO t P PO P A AO A GesLängenveränderungDeltaellsim Der Aluminiumdraht eigentlich ein langer/hoher Zylinder hat anfänglich SolQtyVAX*lXcubicmeter V Aell_ A l V Volumen und damit SolQtymrX*VXkg m rho V rho Aell_ r A l m Masse. Führt man ihm SolQtyQPX*tXJ E Q Pt P t Q Wärmeenergie zu wird seine Temperatur um SolQtydTQX/cX/mXcelsius Deltatheta fracQcm fracPtc rho Aell_ fracP tc r A l dT ansteigen womit er sich SolQtydllX*alpX*dTXm Deltaell ell_ alpha Deltatheta ell_ alpha fracPtc rho Aell_ fracalpha t Prho c A fracalp t Pr c A dl approx dlS dlP- verlängern wird. Deltaell fracalpha t Prho c A dlP-
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