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>> No.11569434 [View]
File: 134 KB, 600x635, 8b773db741ce8b1140022a3a06545403.jpg [View same] [iqdb] [saucenao] [google]
11569434

>>11569182
No, it looks right. The idea of the IVT is that supposing [math] f(c)<f(a) [/math], then there must be an [math] x\in(a,c) [/math] such that [math] f(c)<f(x)<f(a) [/math], which means the function is decreasing. But the function was increasing by hypothesis. So either we have to accept [math] c<a [/math] or that f is decreasing. Either of these is a contradiction.
>>11569210
Water has specific heat capacity [math] c [/math], which you can assume is constant. The thermal energy lost by the 100 ml is the same energy gained by the 300 ml. [math]m_1c\Delta T_1=-m_2c\Delta T_2\implies \Delta T_1=-3\Delta T_2\implies T-80^\circ=-3(T-40^\circ) [/math]. So final temperature is 50 degrees. Yep.
>>11567586
What's the region you are integrating over?
>>11568944
Numerically
>>11567213
Did you have a question?

>> No.11320322 [View]
File: 134 KB, 600x635, 8b773db741ce8b1140022a3a06545403.jpg [View same] [iqdb] [saucenao] [google]
11320322

>>11320003
>bernoulli
No, you can't use Bernoulli in the case of non-steady flow. For a problem like this you could take the computational fluid dynamic (CFD) approach, which is hard, or you could use something based and redpilled called the Buckingham Pi theorem.
First, think about what variables the time to empty the pipe would depend on. I can come up with six: time [math] t [/math] depends on the length of the pipe [math] \ell [/math], area of the pipe [math] A [/math], hydraulic diameter (to account for more x-sections than a circular pipe) [math] D [/math], gravity [math] g [/math], viscosity [math] \mu [/math], and density of fluid [math] \rho [/math]. So we can say there is some function [math] f(t,\ell,A,D,g,\mu,\rho)=0 [/math]. Our goal is to non-dimensionalize this function. We know that [math] f [/math] depends on 3 fundamental dimensions: length, mass, and time. The Buckingham Pi theorem says we can therefore develop a new function of 7-3=4 dimensionless terms called pi terms.
[eqn] F(\pi_1,...,\pi_4)=0\implies\pi_1=G(\pi_2,...,\pi_4) [/eqn]
When you go through the algebra you get
[eqn] \frac{t\mu}{\rho\ell^2}=G\Bigg(\frac{A}{\ell^2},\frac{D}{\ell},\frac{g\ell^3\rho^2}{\mu^2}\Bigg) [/eqn]
for which at this point you would do some experimentation to find exactly what this function[math] G [/math] is. Once you get that, you can find time for any shape and length of pipe for whatever fluid you like.

>> No.11225482 [View]
File: 134 KB, 600x635, 8b773db741ce8b1140022a3a06545403.jpg [View same] [iqdb] [saucenao] [google]
11225482

Math questions:
>>11214727
>>11217334
>>11225374
>>11225391

Physics questions:
>>11213159

Chemistry questions:
>>11218694 (bond formation doesn't always release heat. enthalpy of reaction can be negative bby)

Biology and physiology questions:
>>11211671
>>11214681
>>11214769
>>11218331
>>11222696 (hope ur okay now)

Stupid questions/Use the sticky questions:
>>11213895 (no)
>>11214834
>>11215128 (LaTeX is ugly)
>>11216528 (that's not what superposition is)
>>11219357
>>11219803
>>11222483 (wikipedia)
>>11224306 (I'm interested in this)
>>11224376

lmk if i missed one

>>11225467
oof, no. the theory of boundary layers can rim me.

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