EK physics lecture 5 test

[AG22]

Hey everyone, maybe one of you can shed some light on this:

 

A water tower is filled with water to a depth of 15 m. If a leak forms 10 m above the base of the tower, what will be the velocity of the water as it escapes through the leak?

 

I did this: v = sqrt(2gX10m)

 

However, they solved it using h = 5m, since they reason that the leak happens 5 m below the surface.

 

Now correct me if I'm wrong, but I was always under the impression that height is measured relative to some reference point, usually the base of the container, which is arbitrarily assigned a value of 0m.

 

Gracias!!

[fluorescein]

Hey everyone, maybe one of you can shed some light on this:

 

A water tower is filled with water to a depth of 15 m. If a leak forms 10 m above the base of the tower, what will be the velocity of the water as it escapes through the leak?

 

I did this: v = sqrt(2gX10m)

 

However, they solved it using h = 5m, since they reason that the leak happens 5 m below the surface.

 

Now correct me if I'm wrong, but I was always under the impression that height is measured relative to some reference point, usually the base of the container, which is arbitrarily assigned a value of 0m.

 

Gracias!!

 

The post below mine makes more sense . I'm still curious what page this is on though... and if its in the 7th ed?

[ZW87]

Hey,

 

I think it would be helpful for this problem to draw a picture. The tower is 15m tall above ground and the leak is towards the top at a height of 10m. That means that only 5m of water will leak out. Therefore, you only have 5m of water with a velocity, as the water in the lower 10m will just remain in the tower.

 

Hope this makes sense.

[ontariostudent]

If the leak is not at the base, anything below the leak has no effect on the velocity of the water. Imagine the water below the leak as a flat surface (eg. ground) and it will make more sense. Nothing below the leak will escape from the tower, so it is essentially the new base.

[xenox2]

to fully appreciate the concept behind the question you have to solve this using bernoulli's equation P + pgh + 0.5pv² = constant in a system

 

P + pgh + 0.5pv² = P + pgh + 0.5pv²

atmospheric pressure P cancels out since it is the same at water surface and the leak

pgh + 0.5pv² = pgh + 0.5pv²

if we use reference point h = 0 at the point of leak, then h = 5 at water surface, and since velocity is negligible at water surface, therefore 0.5pv² = 0 at water surface and pgh = 0 at h=0

the equation becomes pgh = 0.5pv² where h = 5

 

 

suppose you want to use reference point h=0 at the bottom of the container,

pgh + 0.5pv² = pgh + 0.5pv²

water surface velocity still equals zero, but you can no longer cancel out pgh from one side, and you would get

pgh = pgh + 0.5pv²

at the water surface h = 15

at the leak h = 10

pg(15) = pg(10) + 0.5pv², and simple algebra brings you back to

pgh = 0.5pv² where h = 5

[AG22]

Hey thanks for the responses everyone! That totally makes sense. Much appreciated

 

And fluorescein it's on p 188 q 102 7th ed.