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http://en.wikipedia.org/wiki/Particle_size | # Particle size
Particle size is a notion introduced for comparing dimensions of solid particles (flecks), liquid particles (droplets), or gaseous particles (bubbles).
The notion of particle size applies to
The particle size of a spherical object can be unambiguously and quantitatively defined by its diameter.
However, a typical material object is likely to be irregular in shape and non-spherical. The above quantitative definition of particle size cannot be applied to non-spherical particles. There are several ways of extending the above quantitative definition, so that a definition is obtained that also applies to non-spherical particles. Existing definitions are based on replacing a given particle with an imaginary sphere that has one of the properties identical with the particle.
• Volume based particle size equals the diameter of the sphere that has same volume as a given particle.
$D = 2 \sqrt[3] {\frac{3V}{4\pi}}$
where
$D$: diameter of representative sphere
$V$: volume of particle
• Weight based particle size equals the diameter of the sphere that has same weight as a given particle.
$D = 2 \sqrt[3] {\frac{3W}{4\pi dg}}$
where
$D$: diameter of representative sphere
$W$: weight of particle
$d$: density of particle
$g$: gravitational constant
• Area based particle size equals the diameter of the sphere that has the same surface area as a given particle.
$D = 2 \sqrt[2] {\frac{A}{4\pi}}$
where
$D$: diameter of representative sphere
$A$: surface area of particle
Another complexity in defining particle size appears for particles with sizes below a micrometre. When particle becomes that small, thickness of interface layer becomes comparable with the particle size. As a result, position of the particle surface becomes uncertain. There is convention for placing this imaginary surface at certain position suggested by Gibbs and presented in many books on Interface and Colloid Science.[1][2][3][4][5][6]
Definition of the particle size for an ensemble (collection) of particles presents another problem. Real systems are practically always polydisperse, which means that the particles in an ensemble have different sizes. The notion of particle size distribution reflects this polydispersity. There is often a need of a certain average particle size for the ensemble of particles. There are several different ways of defining such a particle size.
• There is an International Standard on presenting various characteristic particle sizes.[7] This set of various average sizes includes median size, geometric mean size, average size.
There are several methods for measuring particle size. Some of them are based on light, other on ultrasound, or electric field, or gravity, or centrifugation. They are briefly described in the section particle size distribution. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8864879012107849, "perplexity": 1008.5890378058072}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997888283.14/warc/CC-MAIN-20140722025808-00237-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://spmchemistry.blog.onlinetuition.com.my/2012/07/conversion-of-the-unit-of-concentration.html | # Conversion of the Unit of Concentration
## Conversion of the Unit of Concentration
1. The chart above shows how to convert the units of concentration from g dm-3 to mol dm-3 and vice versa.
2. The molar mass of the solute is equal to the relative molecular mass of the solute.
Example 1:
The concentration of a Potassium chloride solution is 14.9 g dm-3. What is the molarity ( mol dm-3) of the solution? [ Relative Atomic Mass: Cl = 35.5; K = 39 ]
Relative Formula Mass of Potassium Chloride (KCl)
= 39 + 35.5 = 74.5
Molar Mass of Potassium Chloride = 74.5 g/mol
Molarity of Potassium Chloride
Molarity = Concentration Molar Mass = 14.9gd m −3 74.5gmol−1 =0.2mol/dm3
Example 2
A solution of barium hydrokxide have molarity 0.1 mol dm-3. What is the concentration of the solution in g dm-3? [Relative Atomic Mass: Ba = 137; O = 16; H = 1 ] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9127719402313232, "perplexity": 4996.73930768253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00219.warc.gz"} |
https://www.physicsforums.com/threads/dervative-of-volume-of-sphere.649429/ | # Dervative of volume of sphere?
1. Nov 4, 2012
### Psyguy22
If you take the derivative of the area of a circle, you get the formula for circumference. When you take the derivative of the volume of the sphere, you do not get the formula for the area of a circle. Why not?
d/dr (4/3pi r^3) =4pi r^2
d/dr (pi r^2)= 2pi r
2. Nov 4, 2012
### Mute
What you actually get when you take the derivative of the volume with respect to radius is the surface area of the ball. Note that there is a technical difference between a ball and a "sphere": a sphere is, strictly speaker, the surface of a ball. It does not include the volume it encloses, whereas a "ball" includes the surface and the volume contained within. Occasionally the forum gets questions about the "volume of a sphere" and someone will answer that it is zero, which is technically correct because the volume of a "sphere", interpreted literally, refers to the volume the surface itself, which is zero, and not the volume contained by the surface; the OP in these cases pretty much always means the volume of the ball and was just unaware of the precise distinction in the terminology. So, just pointing that out.
Anyways, a way to see why it works like this is to consider the following: to build a circle of area $\pi R^2$, you can think of the process of building "shells" of circles of increasing radius, where each shell has an infinitesimal thickness $dr$. By adding more and more shells you are increasing the area of the circle you are building. If you have a circle of radius r, the infinitesimal change in area you get when adding another shell is $dA = 2\pi r dr$ - the circumference of the shell times the thickness.
When you then go and start building a ball in a similar manner, you are not adding shells of circles. Rather, you are adding shells of spheres of thickness $dr$ and surface area $4\pi r^2$. So, the infinitesimal change in volume as you add a shell to a ball of radius r is $dV = 4\pi r^2 dr$ - the surface area times the thickness.
Does that make sense?
3. Nov 4, 2012
### lurflurf
area of circle:circumference::volume of sphere:surface area of sphere
pi r^2:2pi r::4pi r^3/2:4pi r^2
A:A'::V=V'
you get surface area of the sphere
This follows from Stokes theorem
$$\int_\Omega \mathrm {d}\omega = \int_ {\partial \Omega} \omega$$
Last edited: Nov 4, 2012
4. Nov 4, 2012
### Psyguy22
So my old geometry book lied? Isn't a sphere a 3-D shape? Meaning it would have volume? What exactly does the equation 4/3pi r^3 represent then?
5. Nov 4, 2012
### pwsnafu
Sort of. In natural English sphere is a 3D object. In mathematics it's a 2D object. From Wikipedia:
Your geometry textbook wasn't being careful.
Mute already told you: the term is ball.
6. Nov 4, 2012
### Psyguy22
Ok. Thank you. And sorry for the misuse of words. So just to make sure I understand
4/3 pi r^3 is the volume of a 'ball' and 4 pi r^2 is the surface area of a 'sphere'?
7. Nov 4, 2012
### arildno
It is, of course, the surface area of the ball itself, the sphere being its SURROUNDING BOUNDARY.
(Just as the circumference of the DISK constitutes the surrounding circle bounding the disk)
For nice geometrical objects, the surrounding boundary of the object is of 1 dimension lower than the object itself. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.907912015914917, "perplexity": 615.1197623507875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267861899.65/warc/CC-MAIN-20180619041206-20180619061206-00234.warc.gz"} |
https://perminc.com/resources/publications/role-of-medium-heterogeneity-and-viscosity-contrast-in-miscible-flow-regimes-and-mixing-zone-growth-a-computational-pore-scale-approach/ | ## Role of Medium Heterogeneity and Viscosity Contrast in Miscible Flow Regimes and Mixing Zone Growth: A Computational Pore-Scale Approach
Afshari, S., Kantzas, A., Hejazi, H.
DOI: 10.1103/PhysRevFluids.3.054501
Physical Review Fluids, 3(5), May 2018.
## ABSTRACT
Miscible displacement of fluids in porous media is often characterized by the scaling of the mixing zone length with displacement time. Depending on the viscosity contrast of fluids, the scaling law varies between the square root relationship, a sign for dispersive transport regime during stable displacement, and the linear relationship, which represents the viscous fingering regime during an unstable displacement. The presence of heterogeneities in a porous medium significantly affects the scaling behavior of the mixing length as it interacts with the viscosity contrast to control the mixing of fluids in the pore space. In this study, the dynamics of the flow and transport during both unit and adverse viscosity ratio miscible displacements are investigated in heterogeneous packings of circular grains using pore-scale numerical simulations. The pore-scale heterogeneity level is characterized by the variations of the grain diameter and velocity field. The growth of mixing length is employed to identify the nature of the miscible transport regime at different viscosity ratios and heterogeneity levels. It is shown that as the viscosity ratio increases to higher adverse values, the scaling law of mixing length gradually shifts from dispersive to fingering nature up to a certain viscosity ratio and remains almost the same afterwards. In heterogeneous media, the mixing length scaling law is observed to be generally governed by the variations of the velocity field rather than the grain size. Furthermore, the normalization of mixing length temporal plots with respect to the governing parameters of viscosity ratio, heterogeneity, medium length, and medium aspect ratio is performed. The results indicate that mixing length scales exponentially with log-viscosity ratio and grain size standard deviation while the impact of aspect ratio is insignificant. For stable flows, mixing length scales with the square root of medium length, whereas it changes linearly with length during unstable flows. This scaling procedure allows us to describe the temporal variation of mixing length using a generalized curve for various combinations of the flow conditions and porous medium properties.
A full version of this paper is available on ResearchGate Online. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212046265602112, "perplexity": 1161.1990471668341}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337731.82/warc/CC-MAIN-20221006061224-20221006091224-00030.warc.gz"} |
https://cse331.wordpress.com/2009/12/04/lect-36-weighted-interval-scheduling/ | Posted by: atri | December 4, 2009
## Lect 36: Weighted Interval Scheduling
(Guest post by Santosh Thapa)
Prof Atri started lecture with some recap about weighted interval schedule. Here, he discussed that all jobs are sorted by their finish time and all inputs for weighted interval schedule have non-negative value and output will be a set of schedules which is optimal and our goal is to have maximum optimal solution.
• Input: $n$jobs, $i$th job $(s_i,f_i,v_i)$ which stand for start time, finish time, and value respectively, sorted according to finishing time.
• Output: A schedule $S\subseteq \{1,\dots,n\}$ where for $i\neq j\in S$$i$ and $aj$ re not conflicting.
• Goal: max $\sum_{i\in S} v_i$
Recall we proved last time that
$OPT(j) = max \{ v_j + OPT (p(j)), OPT (j-1)\}$
We could have two option for $j$, it could be in $\mathcal{O}_j$ and it could not be in $\mathcal{O}_j$. And it can be done in linear time.
Question: How can we figure out if $j$ in optimal solution or not (given the values $OPT(1),\dots,OPT(j-1)$)?
Answer: Once we know the value of all optimal solutions, we could compare $v_j+OPT(p(j))$ to $OPT(j-1)$ and if the first term is greater then $j$ is in a optimal schedule.
Professor mentioned that Dynamic programming is similar to Divide and Conquer, where both use recursion but Dynamic programming uses different kind of recursion. In other words, dynamic programming is smarter about solving recursive sub problems.
To compute $OPT(j)$, he discussed about its algorithm using property of optimality, which is
$Compute-OPT( j)$
1. If $j = 0$, return $0$
2. return $\max \{ v_j + Compute-Opt( p(j) ), Compute-Opt( j-1 ) \}$
To prove this algorithm he started proving Lemma.
Lemma: For every $0 \le j \le n$, $OPT(j) = Compute-OPT(j)$.
Proof Idea: Induction on $j$
Base Case: if $j =0$, $Compute- OPT(0) = 0 = OPT(0)$.
Inductive Hypothesis: For $i < j$, $OPT(i) = Compute-OPT(i)$
Inductive Step: Since, $j > 0$ return value is $\max\{ v_j + Compute-OPT(P(j)), Compute-OPT(i)\}$. By the IH, $Compute-OPT(p(j))=OPT(p(j))$ and $Compute-Opt(j-1)=OPT(j-1)$, the proof follows from the recursive formula for $OPT(j)$.
Running time: Exponential
Example:
Here for each job value, it need to call recursively. So, it is exponentially recursive. But important aspect of this kind of algorithm will be helpful in while doing re-computation, which can be done very quickly by storing the recursive value. Even, though it takes more memory, but it will help in solving problem faster.
At the end of the lecture, he wrote the algorithm. We will need a global array $M[j] = null$ or $OPT(j)$.
$MCompOPT(j)$
1. If $j = 0$, return $0$
2. If $M[j] \neq null$, return $M[j]$
3. $M[j] = max\{v_j + MCompOpt(p(j)), MCompOPT(j-1)\}$
4. Return $M[j]$
End of the lecture. Continue of Dec 4th. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 43, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7502875924110413, "perplexity": 17982.727273426142}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806771.56/warc/CC-MAIN-20171123104442-20171123124442-00002.warc.gz"} |
https://dsp.stackexchange.com/questions/8346/fft-function-isnt-working-on-imported-signals-greater-than-a-second | # FFT function isn't working on imported signals greater than a second
I have an fft function in matlab/octave that works great on an imported mono wav/audio signal that is 1 second long, but when the signal imported is more than 1 second the frequency values become incorrect. To check this I created a 2 second signal using audacity the first half of the 2 second signal (1 second of it) was 500hz at 0.3 amplitude and the second half of the 2 second signal (1 second of it) was 200hz at 0.8 amplitude. I also checked it with audacity's analyse spectrum option.
When I use the function on this 2 second signal it shows that the max frequency is at 999.95hz when it should show the max frequency is at 200hz. Can anyone help me get this function to work with signals greater than just 1 second.
I've included the function along with the code that calls the function.
The FUNCTION:
function [freq,amp,ampinv,phase,phase_radtodeg,phase_degtorad,phase_adjindeg,phase_adjinrad,sigfft,sigifft,sigphase]=rtfftphase(vp_sig_orig,Fs,phase_goal)
vp_sig_orig = vp_sig_orig - mean(vp_sig_orig); %remove the mean of your signal due to dc offset
vp_sig_orig=vp_sig_orig';
vp_sig_len=length(vp_sig_orig); %get sample rate from vp fs_rate needs to be an even number?
% Use next highest power of 2 greater than or equal to length(x) to calculate FFT.
nfft= 2^(nextpow2(length(vp_sig_orig)));
% Take fft, padding with zeros so that length(fftx) is equal to nfft
fftx = fft(vp_sig_orig,nfft);
sigfft= fft(vp_sig_orig);
sigifft=ifft(sigfft);
sigphase = unwrap(angle(sigfft')); %get phase of orginal signal
% Calculate the number of unique points
NumUniquePts = ceil((nfft+1)/2);
% FFT is symmetric, throw away second half
fftx = fftx(1:NumUniquePts);
% Take the magnitude of fft of x and scale the fft so that it is not a function of the length of x
mx = abs(fftx)/length(vp_sig_orig); %replaced for testing from stackexchage
% Since we dropped half the FFT, we multiply mx by 2 to keep the same energy.
% The DC component and Nyquist component, if it exists, are unique and should not be multiplied by 2.
if rem(nfft, 2) % odd nfft excludes Nyquist point
mx(2:end) = mx(2:end)*2;
else
mx(2:end -1) = mx(2:end -1)*2;
end
%yamp=(mx(1,:)/max(abs(mx(1,:)))*1); % keep at 1, dont use in function use in script file directly
amp=mx;
ampinv=abs(amp-max(amp));
% This is an evenly spaced frequency vector with NumUniquePts points.
freq_vect = (0:NumUniquePts-1)*vp_sig_len/nfft;
freq=(freq_vect'); %take of round if you want exact numbers
%get phase of new signal
phase = unwrap(angle(fftx)); %get phase of orginal signal
%phase stuff phase adj stuff not fully done yet I don't know the correct way to use it leave at 0 for now
Example: that calls the function
imported_sig_L=imported_sig(:,1)';
%Sort Array Section
array1_sort_amp_norm=sortrows(array1,-2); %sort by amplitude
array1_sort_amp_inv=sortrows(array1,-3); %sort by amplitude inverse
%get max freq checking to make sure freq isn't zero if so go to next row
if (array1_sort_amp_norm(1,1)>0); % To get max amplitude
maxfreq_normA=array1_sort_amp_norm(1,1)
elseif (array1_sort_amp_norm(1,1)<=0); %
maxfreq_normA=array1_sort_amp_norm(2,1)
end
if (array1_sort_amp_inv(1,1)>0); % To get inverted max amplitude
maxfreq_invA=array1_sort_amp_inv(1,1)
elseif (array1_sort_amp_inv(1,1)<=0); %
maxfreq_invA=array1_sort_amp_inv(2,1)
end
Thanks
• I think that you forgot to divide frequency to sample length. – Eddy_Em Mar 23 '13 at 6:42
• @Eddy_Em I tried your suggestion and did freq=(freq_vect')./length(vp_sig_orig) unfortunately it didn't work instead of the freq being 200hz it was 0.0045319hz any other ideas? – Rick T Mar 23 '13 at 7:00
• @Eddy_Em I thought the lines nfft= 2^(nextpow2(length(vp_sig_orig))); and freq_vect = (0:NumUniquePts-1)*vp_sig_len/nfft; would fix this I tried your suggestion and did freq=(freq_vect')./length(vp_sig_orig) unfortunately it didn't work instead of the freq being 200hz it was 0.0045319hz any other ideas? – Rick T Mar 23 '13 at 7:46
• Look into my answer. Let's say your initial data have $N$ points and its length is $T$ seconds, then difference in neighbour points by T-axe would be $\Delta T = T / N$. In frequency domain you will have the same $N$ points, but the max frequency would be equal to $1 / \Delta T$ i.e. $N / T$. So, you just should divide $n$ (number of point of FFT) to $T$ (time length of data sample). – Eddy_Em Mar 23 '13 at 7:54
Here a short example of how to deal with FFT in octave/matlab:
function freq(npts, len, basefreq)
x = [0 : npts - 1] / npts * len;
y = sin(basefreq * x * 2 * pi);
F = abs(fft(y));
half = npts / 2;
F = F(1 : half);
plot([0 : half - 1] / len, F);
printf("Base frequency = %g\n", (find(F == max(F)) - 1) / len);
endfunction
try this function with different parameters.
Here npts is size of data vector; len - time length of data (in seconds); basefreq - base frequency for sine function. First we fit a time axe (x), after that we calculate sinusoid with given frequency, get modulus of its FFT, trim it to half (BTW, another variant is not to trim, but to move frequency zero to middle of fftshift'ed data). After all we plot result of FFT and print out founded base frequency.
• Thanks this helped a lot. I just need to alter one line freq=(freq_vect').*Fs/vp_sig_len; – Rick T Mar 23 '13 at 22:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5417370200157166, "perplexity": 2773.5541260228442}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655900614.47/warc/CC-MAIN-20200709162634-20200709192634-00061.warc.gz"} |
http://worldwidescience.org/topicpages/b/binding+competitive.html | "#### Sample records for binding competitive\n\n1. Paracetamol and cytarabine binding competition in(...TRUNCATED) | "{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, (...TRUNCATED) |
"https://forum.allaboutcircuits.com/threads/high-level-questions-about-adc-dac-codecs-and-hi-end-aud(...TRUNCATED) | "#### Dolmetscher007\n\nJoined Mar 21, 2019\n29\nI am really into audio engineering and digital home(...TRUNCATED) | "{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, (...TRUNCATED) |
https://chemistry.stackexchange.com/questions/64481/schweizers-reagent-from-with-cuo | "# Schweizer's reagent from/with CuO?\n\nWhile attempting to make Schweizer's reagent in household c(...TRUNCATED) | "{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, (...TRUNCATED) |
https://codegolf.stackexchange.com/questions/85847/calculate-the-super-logarithm/86249 | "# Calculate the Super-Logarithm\n\nThis should be a simple challenge.\n\nGiven a number n >= 0, out(...TRUNCATED) | "{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, (...TRUNCATED) |
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