Fundamental quantities and derived quantities

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Length is directly proportional to time in the equation L=Velocity x time, raising questions about its classification as a fundamental quantity. The discussion emphasizes that this classification is based on conventions rather than strict correctness. The seven fundamental physical quantities are not incorrect; they are simply defined according to established conventions. The choice of fundamental quantities depends on the system of units being used. Ultimately, the definitions and conventions shape our understanding of these quantities.
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Homework Statement
If there are only 7 fundamental quantities then how come we can choose the fundamental quantities according to our will like if we take time and speed as fundamental quantities then length is a derived quantity but it's one of the 7 physical quantity so is this stuff working?
Relevant Equations
The seven fundamental physical quantities are length, mass, time, electric current, temperature, amount of substance and intensity of light
L=Velocity x time here length is directly proportional to time so come is it independent as it should be since it's a fundamental physical quantity
 
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It is just a matter of convention. You can choose to work in units where there is only a single fundamental physical dimension. It depends on how you define your system of units.
 
then the 7 fundamental physical quantities that we read are they incorrect ?
 
No. There is no correct/incorrect here. Just definitions.
 
Orodruin said:
No. There is no correct/incorrect here. Just definitions.
so they have been chosen according to what conventions
 
Manish_529 said:
so they have been chosen according to what conventions
Choosing them is the convention.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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