R = v A (look at a drawing of what you have here)
now the way you are supposed to do it
v = L/T
A = L^2
R = L^3/T = (L/T)L^2
so
R = L A/T
but L/T = v
so
R = v A
A horizontal pipe of uniform cross sectional area A empties into a bucket, �filling it at a rate R (unit volume per unit time).
Use dimensional analysis to determine the speed,v, of the fluid in the pipe to within a dimensionless multiplicative constant of order 1.
What happens to v as A decreases (for fi�xed R)?
I understand how dimensional analysis works, however when it says "to within a dimensionless multiplicative constant of order 1" i don't understand what its asking for.
In class we were shown that you change distance=L, Mass=M, Time=T.
Where Rate=L^3/T.
2 answers
well in this case the result is exact, but in general there could be a constant of proportionality between the output and the input that we could not get from dimensional analysis
for example Force on a plate of area A in a fluid of density rho in kg/meter^3 and speed v in L/T
F = ma --> M L/T^2
= L^3 ( M/L^3) L /T^2
= ( M/L^3)L^2 L^2/T^2
= density * Area * v^2
but obviously the actual number will depend on the shape f the object
so
F = some shape constant * density * area * v^2
for example Force on a plate of area A in a fluid of density rho in kg/meter^3 and speed v in L/T
F = ma --> M L/T^2
= L^3 ( M/L^3) L /T^2
= ( M/L^3)L^2 L^2/T^2
= density * Area * v^2
but obviously the actual number will depend on the shape f the object
so
F = some shape constant * density * area * v^2