Question

Water flows at a rate of 8.0 L/s through an opening at the bottom of a tank that has a diameter of 4.0 cm. How high is the water level of this open- top tank?

Answer 1

I will be happy to examine your work.

Answer 2

vf^2=vi^2 + 2ay
(vf^2 - vi^2)/2a=y
8.0 L/s= 0.008 m^3/s
((0.008 m^3/s)^2)/2(9.8)=3.26E-6 m

Answer 3

I haven't even checked the math, but a water level of 3.26 micrometers? That's barely a film, let alone a depth.

give me a break! When you get an answer, at least do a sanity check.

Answer 4

what happened to the density of water?

velocity out= volumeflow/area
= 8L/sec*1m^3/1000L*1/PI(.02^2)=.008/PI*4E-4 m^3/sec
velocity=.002*E4/PI m/sec
velocity=20/pi m/s
Now try your work. The depth will be more than a micro meter.