Answer:
The answer is 1.87nm/s.
Explanation:
The [tex]110g/hr[/tex] water loss must be replaced by [tex]110g/hr[/tex] of sap. 110g of sap corresponds to a volume of
[tex]110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3[/tex]
thus rate of sap replacement is
[tex]106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s[/tex]
The volume of sap in the vessel of length [tex]x[/tex] is
[tex]V = Ax[/tex],
where [tex]A[/tex] is the cross sectional area of the vessel.
For 2000 such vessels, the volume is
[tex]V = 2000Ax[/tex]
taking the derivative of both sides we get:
[tex]\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}[/tex]
on the left-hand-side [tex]\dfrac{dx}{dt}[/tex] is the velocity [tex]v[/tex] of the sap, and on right-hand-side [tex]\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s[/tex]; therefore,
[tex]2.94*10^{-8}m^3/s=2000Av[/tex]
and since the cross-sectional area is
[tex]A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2[/tex];
therefore,
[tex]2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v[/tex]
solving for [tex]v[/tex] we get:
[tex]v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}[/tex]
[tex]\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}[/tex]
which is the upward speed of the sap in each vessel.
