The King's Crown
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:
The following formulas must be used to solve the exercise:
Exercise:
A king has a new crown made for him. For cost reasons the goldsmith uses a gold-silver mixture and processes VO metal. The finished crown rests on the king's head with mO. How much mass gold rgO and how much mass silver rsO did the smith use?
Solution:
If you mix up gold and silver their masses and volumes add up i.e. m m_ + m_ and VV_+V_. Substituting the volumes through mass and density one gets the following two s: tcbhighmathm m_ + m_ quad quad Vfracm_rho_+fracm_rho_ This system of s can be solved using the substitution method; In the first step you solve one it doesn't matter which one we use the first one because it looks easier for one unknown no matter which one: alm_ m-m_ Substitute what you got o the other : alVfracm_rho_+fracm-m_rho_ This can now be solved for the unknown m_; The rules of linear ondimensional algebra must be applied: V fracm_rho_+fracm-m_rho_ rho_rho_ V m_rho_ + rho_m-m_ rho_rho_ V m_rho_ + rho_m-rho_m_ rho_rho_ V - rho_m m_rho_-rho_ m_ fracrho_rho_ V - rho_mrho_-rho_ SolQtymgrgX*rsX*VX-rgX*mX/rsX-rgXkg SolQtymsmX-mgXkg m_ fracrho_rho_ V-rho_ mrho_-rho_ mg m_ m-m_ ms
A king has a new crown made for him. For cost reasons the goldsmith uses a gold-silver mixture and processes VO metal. The finished crown rests on the king's head with mO. How much mass gold rgO and how much mass silver rsO did the smith use?
Solution:
If you mix up gold and silver their masses and volumes add up i.e. m m_ + m_ and VV_+V_. Substituting the volumes through mass and density one gets the following two s: tcbhighmathm m_ + m_ quad quad Vfracm_rho_+fracm_rho_ This system of s can be solved using the substitution method; In the first step you solve one it doesn't matter which one we use the first one because it looks easier for one unknown no matter which one: alm_ m-m_ Substitute what you got o the other : alVfracm_rho_+fracm-m_rho_ This can now be solved for the unknown m_; The rules of linear ondimensional algebra must be applied: V fracm_rho_+fracm-m_rho_ rho_rho_ V m_rho_ + rho_m-m_ rho_rho_ V m_rho_ + rho_m-rho_m_ rho_rho_ V - rho_m m_rho_-rho_ m_ fracrho_rho_ V - rho_mrho_-rho_ SolQtymgrgX*rsX*VX-rgX*mX/rsX-rgXkg SolQtymsmX-mgXkg m_ fracrho_rho_ V-rho_ mrho_-rho_ mg m_ m-m_ ms
Meta Information
Exercise:
A king has a new crown made for him. For cost reasons the goldsmith uses a gold-silver mixture and processes VO metal. The finished crown rests on the king's head with mO. How much mass gold rgO and how much mass silver rsO did the smith use?
Solution:
If you mix up gold and silver their masses and volumes add up i.e. m m_ + m_ and VV_+V_. Substituting the volumes through mass and density one gets the following two s: tcbhighmathm m_ + m_ quad quad Vfracm_rho_+fracm_rho_ This system of s can be solved using the substitution method; In the first step you solve one it doesn't matter which one we use the first one because it looks easier for one unknown no matter which one: alm_ m-m_ Substitute what you got o the other : alVfracm_rho_+fracm-m_rho_ This can now be solved for the unknown m_; The rules of linear ondimensional algebra must be applied: V fracm_rho_+fracm-m_rho_ rho_rho_ V m_rho_ + rho_m-m_ rho_rho_ V m_rho_ + rho_m-rho_m_ rho_rho_ V - rho_m m_rho_-rho_ m_ fracrho_rho_ V - rho_mrho_-rho_ SolQtymgrgX*rsX*VX-rgX*mX/rsX-rgXkg SolQtymsmX-mgXkg m_ fracrho_rho_ V-rho_ mrho_-rho_ mg m_ m-m_ ms
A king has a new crown made for him. For cost reasons the goldsmith uses a gold-silver mixture and processes VO metal. The finished crown rests on the king's head with mO. How much mass gold rgO and how much mass silver rsO did the smith use?
Solution:
If you mix up gold and silver their masses and volumes add up i.e. m m_ + m_ and VV_+V_. Substituting the volumes through mass and density one gets the following two s: tcbhighmathm m_ + m_ quad quad Vfracm_rho_+fracm_rho_ This system of s can be solved using the substitution method; In the first step you solve one it doesn't matter which one we use the first one because it looks easier for one unknown no matter which one: alm_ m-m_ Substitute what you got o the other : alVfracm_rho_+fracm-m_rho_ This can now be solved for the unknown m_; The rules of linear ondimensional algebra must be applied: V fracm_rho_+fracm-m_rho_ rho_rho_ V m_rho_ + rho_m-m_ rho_rho_ V m_rho_ + rho_m-rho_m_ rho_rho_ V - rho_m m_rho_-rho_ m_ fracrho_rho_ V - rho_mrho_-rho_ SolQtymgrgX*rsX*VX-rgX*mX/rsX-rgXkg SolQtymsmX-mgXkg m_ fracrho_rho_ V-rho_ mrho_-rho_ mg m_ m-m_ ms
Contained in these collections:
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Dichte 2 by uz
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Krone des Königs by TeXercises
Asked Quantity:
Masse \(m\)
in
Kilogramm \(\rm kg\)
Physical Quantity
Eigenschaft der Materie
Unit
Base?
SI?
Metric?
Coherent?
Imperial?