some of the symbols did not appear correctly, so I am guessing.
The volume of water that has passed at time T is
Volume=int x(t)dt from 0 to T
one half the initial amount is 1/2 Area*Length
so
1/2 Area*2.0=INt 2.0(1-cos(sqrt g/2))t dt
hmmm. Area of the column is not given.So I wonder about the units of x(t). Seeing on the right, it is L, so I guess the problem maker assumed unit area
1=2 int dt -2int cos(sqrt g/2) t)dt
1=2T+2sin (sqrtg/2)T
it is a quadratic in T, so use the quadratic equation to find T. check my thinking, I did most of this ASCII typing and figuring in my head.
A pipe in the shape of a right-angle elbow contains a vertical section and
a horizontal section with a valve at the right angle. Presently the valve
is closed and the vertical section of the pipe contains a column of water
having length L. When the valve is opened, the amount of water that
passes through the valve, as a function of time is given by
x(t) = L {1 − cos (sqrt(g/L)*t)}
for 0 ¡Ü x ¡Ü L, where g = 9.8 meters per seconds squared is the acceleration
due to gravity. If the initial column length is L = 2.0 meters
a.) how long (in seconds) does it take for one-half of the water column to flow
pass the valve?
b.) How long (in seconds) does it take all of the water column to flow pass the
valve?
1 answer
1 answer