Timeline for Am I making a fatal error with this simplification?
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| Aug 11, 2021 at 14:36 | comment | added | Michael Seifert | (Though now that I think about it, the volume average would be another way to do it, and might give the same thing as the time-average. I will edit my answer shortly to reflect this.) | |
| Aug 11, 2021 at 14:34 | comment | added | Michael Seifert | @J.Murray: Note that to get the desired answer you need to take the time-average of a water molecule's speed as it goes through the nozzle. Your description of how to average the speed sounds more like the average of the speed over the distance (which is very very large.) See my answer. | |
| Aug 11, 2021 at 14:28 | history | edited | J. Murray | CC BY-SA 4.0 |
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| Aug 11, 2021 at 14:22 | comment | added | J. Murray | @rdotrose I have edited my answer to address your follow-up. | |
| Aug 11, 2021 at 14:22 | history | edited | J. Murray | CC BY-SA 4.0 |
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| Aug 11, 2021 at 14:08 | comment | added | nasu | Yes, it's true that the volume of the cone is 1/3 of the volume of the cylinder of the same height. But this does not follow from the water flow situation. The speed depends only on the cross sectional area. You can make the nozzle profile any shape (parabolic, hyprbolic, whatewer) but the rate of flow could be the same if the cross section area of the exit end is the same. And besides, you are trying to explain something by relating to something else which is probably less familiar to the students. | |
| Aug 11, 2021 at 14:07 | comment | added | Quantum Mechanic | To find the average, we have to integrate $v_{\mathrm{spout}}$ from the radius of the cylinder $r_0$ to the radius of the hole at the spout, $r_s$. The integral is thus $3\mathrm{cm/s}\int_0^{r_s} \frac{\pi r_0^2}{\pi(r_0-x)^2}dx = 3\mathrm{cm/s}\frac{r_0r_s}{r_0-r_s}$ | |
| Aug 11, 2021 at 14:05 | comment | added | rdotrose | Yes, I tried to address this at the end when I write "This 9cm/s should be considered an average rate. In reality, the water coming out of the tiny tip would be moving faster, but water at the base of the cone where it is the widest would be traveling more slowly." The term average might be used too loosely. But, wouldn't it still be true that after 1 second of flow, the equivalent amount of water in the cylindrical portion of the hose would need to escape the cone shape at the end. And that the cone would have to expel its volume 3 times to match the volume in the cylinder portion? | |
| Aug 11, 2021 at 14:04 | comment | added | Quantum Mechanic | And thus if $A_{\mathrm{spout}}$ actually goes to zero (a perfectly sharp tip with an infinitesimal hole) then the velocity goes to infinity | |
| Aug 11, 2021 at 13:58 | history | answered | J. Murray | CC BY-SA 4.0 |