A value is slowly opened in a pipeline such that the volume flow rate R varies with time according to the relation R = kt (t>0).
Calculate the total volume of water that flows through the value in the first 10 seconds if k=1.3 m^3s^-2
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Just some thought This question has me so confused
this is is a cylindrical shape so v=pi(r^2)h
dv/dt = kt = r
so... ktdx = (k/2)t^2 + C
My other thought was that N=N<0> e^kt
but somehow dn/dt becomes Kt
Calculate the total volume of water that flows through the value in the first 10 seconds if k=1.3 m^3s^-2
******
Just some thought This question has me so confused
this is is a cylindrical shape so v=pi(r^2)h
dv/dt = kt = r
so... ktdx = (k/2)t^2 + C
My other thought was that N=N<0> e^kt
but somehow dn/dt becomes Kt
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I would take the integral of R dt from t= 0 to t = 10;
integral (R dt) = integral (kt dt) = k*t^2 / 2 evaluated from t= 0 to t = 10
= 1.3 * 100 / 2 = 65 m^3.
integral (R dt) = integral (kt dt) = k*t^2 / 2 evaluated from t= 0 to t = 10
= 1.3 * 100 / 2 = 65 m^3.