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strawberrymilkbunny · 26-30, F
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Lol

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3 and 4 at the same time. 1 and 2 will never fill if all containers are open at the top.
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@Volfield: yes man you are right. But the diagram is drawn with rough hands not to pinpoint accuracy. And flow is weak not to introduce turbulence. 🙂🙂
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Pythagorean cup is one of the coolest ways to learn about fluid dynamics and how air pressure affects these kind of systems and a really cool party trick :)
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@Volfield: I studied fluid dynamics during my graduation in mechanical engineering.
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You are right.
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depends on the flow rate from the spigot.. it's going to be 1 or 3
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3
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@samdx008: agreed, if the faucet only drips as depicted.
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4 if by filled you mean up to the top.
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No. The third has to be filled up to the point where it flows into the 4th container. So the 4th will be filled up to the top, then the 3.
If it's just semi to the top then it's the third.
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@GermanAf: oh. You taken the diagram to be exact accurate to the top. The diagram is drawn with rough hands. But your concept is right.
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:)
abe182 · 51-55, M
Depending on a strong water flow it will be 1. A weak flow it'll be 3.
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Laminar flow.
VeronicaPrincess · 61-69
4 - it's upper limit is at the lowest level.
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3rd man.
shuhak · M
3. Gravity will prevent contents of 3 moving rapidly into 4 (anti-siphon). What you have here is the basic works of a toilet.
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Yup. Nope just a general question.
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Right...
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@samdx008: in order for 4 to fill it has to get to the top of the overflow line
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@bigbill: yess.
Doctrble · 51-55, M
4 if you talking filled to top otherwise 1
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No. 3rd.
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Depends how fast the tap is on.
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Wrong. 3
MrBrownstone · 46-50, M
@samdx008: Atleast you edited the question.
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@MrBrownstone: oh. Sorry. I missed to write filled. I got what you understood. You thought which one will get first. I am sorry.
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Lol. Nope.🙂
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@samdx008: Haha. Yeah I scrolled down after putting down my answer and realised its wrong. And I saw why too. :)
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@Aidolovemostofyourthoughts: lol.😁😁🙂
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MasterLee · 56-60, M
Has to be 3
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Yup.
clayanthony · 51-55, M
i'll go for 3
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Right.
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Right...
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Wrong. 3rd.

 
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