Destroy a wine barrel
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
\(\LaTeX\)
Amédée Guillemin, , 1872, illustration, The forces of nature: a popular introduction to the study of physical phenomena p. 69
<Wikipedia> (retrieved on November 28, 2022)
Need help? Yes, please!
The following quantities appear in the problem:
Kraft \(F\) / Druck \(p\) / Fläche \(A\) / Ortsfaktor \(g\) / Höhe \(h\) / Radius \(r\) / Dichte \(\varrho\) /
The following formulas must be used to solve the exercise:
\(p = \dfrac{F}{A} \quad \) \(A = \pi r^2 \quad \) \(p = \varrho g h \quad \)
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Exercise:
In the th century french physicist Blaise Pascal claimed that he could destroy a full wine barrel with a few glasses of wine. A long tube rO inner radius was inserted o a water-filled wine barrel RO lid radius and water was filled o the long tube until the barrel burst. This happened at a filling height of hO. By what factor is the force on the lid bigger than the weight of the liquid in the small tube?
Solution:
Geg r rO r R RO R h h GesfactorN % The pressure of the water column is NewQtyrwkgpmk SolQtyprwX*ncgn*hXPa al p rho g h rw ncg h p. This pressure acts on the lid its circular surface is SolQtyApi*RX^metersquared al A pi R^ pi qtyR^ A The force acting on the circular lid is then SolQtyFpX*AXN al F p A rho g h pi R^ pirho ghR^ p A F. In the tube it has SolQtyVpi*rX^*hXcubicmeter V pi r^ h pi qtyr^ h V water with a mass of SolQtymrwX*VXkg m rho V rw V m and SolQtyFgmX*ncgnN sscFG mg m ncg Fg in weight. The force acting on the lid is therefore SolQtyNFX/FgX N fracFsscFG N times bigger than the weight of the water in the small lid. N N
In the th century french physicist Blaise Pascal claimed that he could destroy a full wine barrel with a few glasses of wine. A long tube rO inner radius was inserted o a water-filled wine barrel RO lid radius and water was filled o the long tube until the barrel burst. This happened at a filling height of hO. By what factor is the force on the lid bigger than the weight of the liquid in the small tube?
Solution:
Geg r rO r R RO R h h GesfactorN % The pressure of the water column is NewQtyrwkgpmk SolQtyprwX*ncgn*hXPa al p rho g h rw ncg h p. This pressure acts on the lid its circular surface is SolQtyApi*RX^metersquared al A pi R^ pi qtyR^ A The force acting on the circular lid is then SolQtyFpX*AXN al F p A rho g h pi R^ pirho ghR^ p A F. In the tube it has SolQtyVpi*rX^*hXcubicmeter V pi r^ h pi qtyr^ h V water with a mass of SolQtymrwX*VXkg m rho V rw V m and SolQtyFgmX*ncgnN sscFG mg m ncg Fg in weight. The force acting on the lid is therefore SolQtyNFX/FgX N fracFsscFG N times bigger than the weight of the water in the small lid. N N
Meta Information
Exercise:
In the th century french physicist Blaise Pascal claimed that he could destroy a full wine barrel with a few glasses of wine. A long tube rO inner radius was inserted o a water-filled wine barrel RO lid radius and water was filled o the long tube until the barrel burst. This happened at a filling height of hO. By what factor is the force on the lid bigger than the weight of the liquid in the small tube?
Solution:
Geg r rO r R RO R h h GesfactorN % The pressure of the water column is NewQtyrwkgpmk SolQtyprwX*ncgn*hXPa al p rho g h rw ncg h p. This pressure acts on the lid its circular surface is SolQtyApi*RX^metersquared al A pi R^ pi qtyR^ A The force acting on the circular lid is then SolQtyFpX*AXN al F p A rho g h pi R^ pirho ghR^ p A F. In the tube it has SolQtyVpi*rX^*hXcubicmeter V pi r^ h pi qtyr^ h V water with a mass of SolQtymrwX*VXkg m rho V rw V m and SolQtyFgmX*ncgnN sscFG mg m ncg Fg in weight. The force acting on the lid is therefore SolQtyNFX/FgX N fracFsscFG N times bigger than the weight of the water in the small lid. N N
In the th century french physicist Blaise Pascal claimed that he could destroy a full wine barrel with a few glasses of wine. A long tube rO inner radius was inserted o a water-filled wine barrel RO lid radius and water was filled o the long tube until the barrel burst. This happened at a filling height of hO. By what factor is the force on the lid bigger than the weight of the liquid in the small tube?
Solution:
Geg r rO r R RO R h h GesfactorN % The pressure of the water column is NewQtyrwkgpmk SolQtyprwX*ncgn*hXPa al p rho g h rw ncg h p. This pressure acts on the lid its circular surface is SolQtyApi*RX^metersquared al A pi R^ pi qtyR^ A The force acting on the circular lid is then SolQtyFpX*AXN al F p A rho g h pi R^ pirho ghR^ p A F. In the tube it has SolQtyVpi*rX^*hXcubicmeter V pi r^ h pi qtyr^ h V water with a mass of SolQtymrwX*VXkg m rho V rw V m and SolQtyFgmX*ncgnN sscFG mg m ncg Fg in weight. The force acting on the lid is therefore SolQtyNFX/FgX N fracFsscFG N times bigger than the weight of the water in the small lid. N N
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Asked Quantity:
Verhältnis / Anteil \(\eta\)
in
Prozentsatz \(\rm \eta\)
Physical Quantity
Unit