http://s8.postimage.org/hfhhlg6r9/untitled12.jpg The large container shown in th

Question

http://s8.postimage.org/hfhhlg6r9/untitled12.jpg

The large container shown in the cross section to the right is filled with
a liquid of density 1.1 × 103 kg/m3. A small hole of area 2.5 × 10-6 m2
is opened in the side of the container a distance h below the liquid
surface, which allows a stream of liquid to flow through the hole and
into a beaker placed to the right of the container. At the same time,
liquid is also added to the container at an appropriate rate so that h
remains constant. The amount of liquid collected in the beaker in 2.0
minutes is 7.2 × 10-4 m3.
(a) Calculate the volume rate of flow of liquid from the hole in m3/s.
(b) Calculate the speed of the liquid as it exits from the hole.
(c) Calculate the height h of liquid needed above the hole to cause the speed your determined in part (b).
(d) Suppose that there is now less liquid in the container so the the height h is reduced to h/2. In relation to
the beaker, where will the liquid hit the table top
____ Left of the beaker ____ In the beaker ____ Right of the beaker
Justify your answer.

Explanation / Answer

1.well the first one kinda of easy you are told that in 2 min you get 7.2 x 10^-4 m^3 so to get that and dived by 60 and you get the m^3/s 2.Equation Q=Av shows how velocity (v) and Flow Rate (Q) are related. Q= (7.2 x 10^-4 m^3/120) from question 1 and as allways A is the area of the hole. 3. i dont understand the question sry 4. if there is less pressure the Flow rate will decrease and so will the Velocity, causing the stream to be weaker and not reach the beaker so it lands between the beaker and the tabletop. there is a proportion to this but i dont want to calculate it

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