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EK physics lecture 5 test


AG22

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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!! :)

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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? :)

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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.

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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

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