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pacelweb
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In my Further Maths class today...
1. Produce a function which can b used to predict the depth of water in a can with a hole in it
The experiment, conducted by a friend and I, gave results of the height left from times 0s to 1010s. At t=0, depth (h) = 0.125m
Torricelli's equation [tex]v=\sqrt{2gh}[/tex]
We can deduce by inspection of a graph plotted of the data that the rate of flow out of the can is proportional to the height left (Note this is a decay, hence the negative):
[tex]\frac{dh}{dt}\propto{-h}[/tex]
[tex]\frac{dh}{dt}=-kt[/tex]
Solving the differential equation we get:
[tex]h=Ae^{-kt}[/tex]
Substituting t=0, A=0.125m
[tex]h=0.125e^{-kt}[/tex]
Attempting to find k, and using autograph, I substituted different values of t and their corresponding values of h from my results. However, I found that t was different in each case- the higher the value of t, the higher value of k; or graphically, using a lower value or k, the function fits the data at low values of t, for higher values of k the function is a better fit at higher t values.
I cannot figure out in my mind what physical property(ies) k represents. There was a hint from my teacher that velocity may play a part, and I have looked at Torricelli's [tex]v=\sqrt{}2gh[/tex] but cannot link into how this would help me.
Could anyone suggest what k may actually represent?
P.S sorry if this should be in Introductory Physics, but as we have done it in our maths lesson I naturally put it in here...
Many thanks, Philip Wiseman
1. Produce a function which can b used to predict the depth of water in a can with a hole in it
The experiment, conducted by a friend and I, gave results of the height left from times 0s to 1010s. At t=0, depth (h) = 0.125m
Homework Equations
Torricelli's equation [tex]v=\sqrt{2gh}[/tex]
The Attempt at a Solution
We can deduce by inspection of a graph plotted of the data that the rate of flow out of the can is proportional to the height left (Note this is a decay, hence the negative):
[tex]\frac{dh}{dt}\propto{-h}[/tex]
[tex]\frac{dh}{dt}=-kt[/tex]
Solving the differential equation we get:
[tex]h=Ae^{-kt}[/tex]
Substituting t=0, A=0.125m
[tex]h=0.125e^{-kt}[/tex]
Attempting to find k, and using autograph, I substituted different values of t and their corresponding values of h from my results. However, I found that t was different in each case- the higher the value of t, the higher value of k; or graphically, using a lower value or k, the function fits the data at low values of t, for higher values of k the function is a better fit at higher t values.
I cannot figure out in my mind what physical property(ies) k represents. There was a hint from my teacher that velocity may play a part, and I have looked at Torricelli's [tex]v=\sqrt{}2gh[/tex] but cannot link into how this would help me.
Could anyone suggest what k may actually represent?
P.S sorry if this should be in Introductory Physics, but as we have done it in our maths lesson I naturally put it in here...
Many thanks, Philip Wiseman
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