Nordhäuser Doppelkorn
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
\(\LaTeX\)
, 2023, digital photograph, Echter Nordhäuser
<https://www.echter-nordhaeuser> (retrieved on February 26, 2023)
Need help? Yes, please!
The following quantities appear in the problem:
Masse \(m\) / Volumen \(V\) / Dichte \(\varrho\) / Verhältnis / Anteil \(\eta\) /
The following formulas must be used to solve the exercise:
\(\eta = \dfrac{a}{A} \quad \) \(\varrho = \dfrac{m}{V} \quad \)
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:
In einer VO-Schnaps-Flasche BrandNameNordhäuser Doppelkorn beträgt die Alkohol-Konzentration .vol-% der andere Teil kann näherungsweise als Wasser betrachtet werden. Wie viel Massenprozent Alkohol sind in dem Schnaps?Formelbuch Die Dichte von Alkohol beträgt rO.
Solution:
newqtyVo.deciliter newqtyVVon cubicmeter newqtyetVo.percent newqtyetVetVon newqtyRWkgpcm newqtyRAo.grampercubiccentimeter newqtyRARAon ekgpcm % Geg V Vo V eta_V etVo etV textWasser rightarrow sscrhoW RW textAlkohol rightarrow sscrhoA RAo RA % GesMassen-Anteileta_mtext in sipercent % Das Volumen des Alkohols beträgt solqtyVAeta_V VetVn*Vncubicmeter al sscVA VAf etVV VA. Seine Masse beträgt somit solqtymAsscrhoAVAfRAn*VAnkg al sscmA sscrhoAsscVA mAf RA VA mA. Das Volumen des als Wasser angenäherten Restes beträgt solqtyVW-eta_V VVn-VAncubicmeter al sscVW V-sscVA V-VAf VWf V-VA VW die Masse davon ist solqtymWsscrhoWVWfRWn*VWnkg al sscmW sscrhoWsscVW mWf RWVW mW. Der Massenanteil des Alkohols ist demnach solqtyetmfraceta_VsscrhoAeta_VsscrhoA + -eta_V sscrhoWmAn/mAn+mWn convtopercentetm al eta_m fracsscmAsscmA+sscmW fracmAfmAf+mWf etmf fracmAmA+mW etm etmCTTTT. % eta_m etmf etmCTT
In einer VO-Schnaps-Flasche BrandNameNordhäuser Doppelkorn beträgt die Alkohol-Konzentration .vol-% der andere Teil kann näherungsweise als Wasser betrachtet werden. Wie viel Massenprozent Alkohol sind in dem Schnaps?Formelbuch Die Dichte von Alkohol beträgt rO.
Solution:
newqtyVo.deciliter newqtyVVon cubicmeter newqtyetVo.percent newqtyetVetVon newqtyRWkgpcm newqtyRAo.grampercubiccentimeter newqtyRARAon ekgpcm % Geg V Vo V eta_V etVo etV textWasser rightarrow sscrhoW RW textAlkohol rightarrow sscrhoA RAo RA % GesMassen-Anteileta_mtext in sipercent % Das Volumen des Alkohols beträgt solqtyVAeta_V VetVn*Vncubicmeter al sscVA VAf etVV VA. Seine Masse beträgt somit solqtymAsscrhoAVAfRAn*VAnkg al sscmA sscrhoAsscVA mAf RA VA mA. Das Volumen des als Wasser angenäherten Restes beträgt solqtyVW-eta_V VVn-VAncubicmeter al sscVW V-sscVA V-VAf VWf V-VA VW die Masse davon ist solqtymWsscrhoWVWfRWn*VWnkg al sscmW sscrhoWsscVW mWf RWVW mW. Der Massenanteil des Alkohols ist demnach solqtyetmfraceta_VsscrhoAeta_VsscrhoA + -eta_V sscrhoWmAn/mAn+mWn convtopercentetm al eta_m fracsscmAsscmA+sscmW fracmAfmAf+mWf etmf fracmAmA+mW etm etmCTTTT. % eta_m etmf etmCTT
Meta Information
Exercise:
In einer VO-Schnaps-Flasche BrandNameNordhäuser Doppelkorn beträgt die Alkohol-Konzentration .vol-% der andere Teil kann näherungsweise als Wasser betrachtet werden. Wie viel Massenprozent Alkohol sind in dem Schnaps?Formelbuch Die Dichte von Alkohol beträgt rO.
Solution:
newqtyVo.deciliter newqtyVVon cubicmeter newqtyetVo.percent newqtyetVetVon newqtyRWkgpcm newqtyRAo.grampercubiccentimeter newqtyRARAon ekgpcm % Geg V Vo V eta_V etVo etV textWasser rightarrow sscrhoW RW textAlkohol rightarrow sscrhoA RAo RA % GesMassen-Anteileta_mtext in sipercent % Das Volumen des Alkohols beträgt solqtyVAeta_V VetVn*Vncubicmeter al sscVA VAf etVV VA. Seine Masse beträgt somit solqtymAsscrhoAVAfRAn*VAnkg al sscmA sscrhoAsscVA mAf RA VA mA. Das Volumen des als Wasser angenäherten Restes beträgt solqtyVW-eta_V VVn-VAncubicmeter al sscVW V-sscVA V-VAf VWf V-VA VW die Masse davon ist solqtymWsscrhoWVWfRWn*VWnkg al sscmW sscrhoWsscVW mWf RWVW mW. Der Massenanteil des Alkohols ist demnach solqtyetmfraceta_VsscrhoAeta_VsscrhoA + -eta_V sscrhoWmAn/mAn+mWn convtopercentetm al eta_m fracsscmAsscmA+sscmW fracmAfmAf+mWf etmf fracmAmA+mW etm etmCTTTT. % eta_m etmf etmCTT
In einer VO-Schnaps-Flasche BrandNameNordhäuser Doppelkorn beträgt die Alkohol-Konzentration .vol-% der andere Teil kann näherungsweise als Wasser betrachtet werden. Wie viel Massenprozent Alkohol sind in dem Schnaps?Formelbuch Die Dichte von Alkohol beträgt rO.
Solution:
newqtyVo.deciliter newqtyVVon cubicmeter newqtyetVo.percent newqtyetVetVon newqtyRWkgpcm newqtyRAo.grampercubiccentimeter newqtyRARAon ekgpcm % Geg V Vo V eta_V etVo etV textWasser rightarrow sscrhoW RW textAlkohol rightarrow sscrhoA RAo RA % GesMassen-Anteileta_mtext in sipercent % Das Volumen des Alkohols beträgt solqtyVAeta_V VetVn*Vncubicmeter al sscVA VAf etVV VA. Seine Masse beträgt somit solqtymAsscrhoAVAfRAn*VAnkg al sscmA sscrhoAsscVA mAf RA VA mA. Das Volumen des als Wasser angenäherten Restes beträgt solqtyVW-eta_V VVn-VAncubicmeter al sscVW V-sscVA V-VAf VWf V-VA VW die Masse davon ist solqtymWsscrhoWVWfRWn*VWnkg al sscmW sscrhoWsscVW mWf RWVW mW. Der Massenanteil des Alkohols ist demnach solqtyetmfraceta_VsscrhoAeta_VsscrhoA + -eta_V sscrhoWmAn/mAn+mWn convtopercentetm al eta_m fracsscmAsscmA+sscmW fracmAfmAf+mWf etmf fracmAmA+mW etm etmCTTTT. % eta_m etmf etmCTT
Contained in these collections:
-
Alkoholgetränk by TeXercises
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Dichte 2 by uz
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Dichte II by pw
Asked Quantity:
Verhältnis / Anteil \(\eta\)
in
Prozentsatz \(\rm \eta\)
Physical Quantity
Unit