text:
# a table of mine won't compile

I have had the problem of compiling my document with the following table in it:

\begin{table}[H]

\begin{center}
\caption{Input parameters associated with the hydraulic model} \label{parametersfinal}
\begin{tabular}{ll|ll}
\textbf{Parameter} & \textbf{Value} & \textbf{Parameter} & \textbf{Value}\\
\hline
$T_{ref}$ [K] & 273,16 & $\kappa_s$ [-] & $0,5 \cdot 10^{-3}$\\
$E_{f,ref}$ [N/m\textsuperscript{2}] & $50 \cdot 10^6$ & $r$ [-] & 0,04 \\
$E_{f,inc}$ [N/m\textsuperscript{2}/K] & $10 \cdot 10^6$ & $\beta$ [(N/m\textsuperscript{2})\textsuperscript{-1}] & $0,08 \cdot 10^{-6}$ \\
$\nu_f$ [-] & 0,45 & $\alpha$ [-] & 9 \\
$G_0$ [N/m\textsuperscript{2}] & $700 \cdot 10^6$ &  $p_{0,ref}$ [N/m\textsuperscript{2}] & -395 $\cdot 10^6$ \\
$\kappa_0$ [-] & 0,01 & $m$ [-] & 1 \\
$p^{*}_c$ [N/m\textsuperscript{2}] & $-40 \cdot 10^3$ & $(p^{*}_{y0})_{in}$ [N/m\textsuperscript{2}] & -500 $\cdot 10^3$ \\
$\lambda_0$ [-] & 0,05 & $Y_{ref}$ [m] & 0\\
$\gamma$ [-] & 1 & $\Delta p^{*}_{y0}$ [N/m\textsuperscript{2}] & 0 \\
$k_t$ [-] & 0,09 & $(S_{c,seg})_{in}$ [N/m\textsuperscript{2}] & $20 \cdot 10^6$  \\
$M$ [-] & 1,11 & $K_w$ [N/m\textsuperscript{2] & $10^9$ \\
\hline
\end{tabular}
\end{center}
\end{table}

I don't get any error messages or anything peculiar in my log, I just click the compile button and it just starts working and keeps at it forever until I abort it. Someone can help me?

• Is it safe to assume that you have a documentclass as well as \begin{document} and \end{document}? – AML May 16 at 13:32
• You're missing a } in the last \textsuperscript. – Torbjørn T. May 16 at 13:33
• Generally speaking, when compilation never stops, it usually means that the compiler is not able to reach the end of the document. In my experience, this also means a missing bracket or a missing ; in a tikz component. – Emilie Picard-Cantin May 16 at 14:40

As Torbjørn T. has already pointed out in a comment, the immediate problem is a missing } character after 2 in the expression [N/m\textsuperscript{2] in the final row of the table.

More generally, you may want to make the table [a lot!] more readable, say, by changing from 4 to 6 columns, so that the units of the parameters can be displayed in separate columns. Do also take the time to learn how to use the machinery of the siunitx package to display scientific units, and numbers, in a more LaTeX-y way, using the package's \si and \num macros.

\documentclass{article}
\usepackage{amsmath,siunitx,caption}
\sisetup{output-decimal-marker = {,},
per-mode              = symbol,
exponent-product      = \cdot,
retain-unity-mantissa = false }
\usepackage{array} % for '\newcolumntype' macro
\newcolumntype{L}{>{$}l<{$}}  % left-alignment, automatic math mode
\begin{document}
\begin{table}
\centering
\caption{Input parameters associated with the hydraulic model} \label{parametersfinal}
\begin{tabular}{@{} Lll | Lll @{}}
$Symbol$ & Units & Value &
$Symbol$ & Units & Value \\
\hline
T_{\textrm{ref}} & \si{\kelvin}$^{\strut}$ & \num{273,16} &
\kappa_s & - & \num{0,5e-3}\\
E_{f,\textrm{ref}} & \si{\newton\per\meter\squared} & \num{50e6} &
r & - & \num{0,04} \\
E_{f,\textrm{inc}} & \si{\newton\per\meter\squared\per\kelvin} & \num{10e6} &
\beta & \si{(\newton\per\meter\squared)\raiseto{-1}} & \num{0,08e-6} \\
\nu_f & - & 0,45 &
\alpha & - & 9 \\
G_0 & \si{\newton\per\meter\squared} & \num{700e6} &
p_{0,\textrm{ref}} & \si{\newton\per\meter\squared} & \num{-395e6} \\
\kappa_0 & - & 0,01 &
m & - & 1 \\
p^{*}_c & \si{\newton\per\meter\squared} & \num{-40e3} &
(p^{*}_{y0})_{\mathrm{in}} & \si{\newton\per\meter\squared} & \num{-500e3} \\
\lambda_0 & - & 0,05 &
Y_{\textrm{ref}} & \si{\meter} & 0\\
\gamma & - & 1 &
\Delta p^{*}_{y0} & \si{\newton\per\meter\squared} & 0 \\
k_t & - & 0,09 &
(S_{c,\textrm{seg}})_{\textrm{in}} & \si{\newton\per\meter\squared} & \num{20e6} \\
M & - & \num{1,11} &
K_w & \si{\newton\per\meter\squared} & \num{1e9} \\
\hline
\end{tabular}
\end{table}
\end{document}

Just for the sake of comparison, here's a screenshot of your original table. In addition to having a more "cramped" look due to the use of just 4 columns, note that there are several typographic inconsistencies (highlighted in yellow) in the right-hand column, relating to either too much whitespace after the decimal marker or the use of a text-dash symbol instead of a mathematical-minus symbol. These inconsistencies are avoided automatically if one uses the \num macro.

• A separate comment: It might make sense, in terms of enhancing the readability of the tabular material, to replace all instances of \si{\newton\per\meter\squared} with \si{\pascal}. – Mico May 16 at 17:33

More or less on the same basis, but I propose to put units in their own column. Don't use the center environment with table: it will add unwanted vertical spacing. Use the \centering command instead. With siunitx you can define your own abbreviations for units.

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{siunitx, caption}
\DeclareSIUnit{\Nsqm}{\N\per\square\m}

\begin{document}

\begin{table}[h]
\centering\setlength{\extrarowheight}{3pt}
\sisetup{per-mode =symbol, exponent-product=\cdot, copy-decimal-marker}
\caption{Input parameters associated with the hydraulic model} \label{parametersfinal}
\begin{tabular}{lll|lll}
\multicolumn{2}{l}{\bfseries Parameter} & \textbf{Value} & \multicolumn{2}{l}{\bfseries Parameter} & \textbf{Value}\\
\hline
$T_\text{ref}$ & [K] & \num{273,16}& $\kappa_s$ & -- & \num{0,5e-3}\\
$E_{f,\text{ref}}$ & [\si{\Nsqm}] & \num{50 e6} & $r$ & -- & \num{0,04} \\
$E_{f,\text{inc}}$ & [\si{{\Nsqm}\per\K}] & \num{10 e6} & $\beta$ & [\si{(\Nsqm)^{-1}}]& \num{0,08e-6}\\
$\nu_f$ & -- & \num{0,45} & $\alpha$ & -- & 9 \\
$G_0$ & [\si{\Nsqm}] & \num{700 e6} & $p_{0, \text{ref}}$ & [\si{\Nsqm}] & \num{-395e6} \\
$\kappa_0$ & -- & 0,01 & \si{m} &– & 1 \\
$p^{*}_c$ & [N/m\textsuperscript{2}] & \num{-40 e3} & $(p^{*}_{y0})_{\text{in}}$ & [\si{\Nsqm}] & \num{-500 e3} \\
$\lambda_0$ & -- & \num{0,05} & $Y_\text{ref}$ & [m] & 0\\
$\gamma$ & -- & 1 & $\Delta p^{*}_{y0}$ & \si{\Nsqm}] & 0 \\
$k_t$ & -- & \num{0,09} & $(S_{c,\text{seg}})_\text{in}$ & [\si{\Nsqm}] & \num{20 e6} \\
$M$ & -- & \num{1,11} & $K_w$ & [\si{\Nsqm}] & \num{e9} \\
\hline
\end{tabular}
\end{table}

\end{document}

a small variation of nice Mico answer. differences are:

• columns with units use s column specifier
• columns with numbers use S column specifier

due to this changes the table code is slightly shorter:

\documentclass{article}
\usepackage{amsmath}
\usepackage{caption}
\usepackage{siunitx}
\sisetup{input-symbols = {( - )},
output-decimal-marker = {,},
exponent-product      = \cdot,
table-unit-alignment  = left
}
\newcommand\mcc[1]{\multicolumn{1}{c}{#1}}
\usepackage{array} % for '\newcolumntype' macro
\newcolumntype{L}{>{$}l<{$}}  % left-alignment, automatic math mode

\begin{document}
\begin{table}[htb]
\centering
\caption{Input parameters associated with the hydraulic model}
\label{parametersfinal}
\renewcommand\arraystretch{1.1}
\begin{tabular}{@{} LsS | LsS @{}}
$Symbol$            &   \mcc{Units}                         & {Value }
& $Symbol$          &   \mcc{Units}                             & {Value}   \\
\hline
T_{\textrm{ref}}    & \kelvin                               & 273,16
& \kappa_s          & {-}                                       & 0,5e-3    \\
E_{f,\textrm{ref}}  & \pascal             & 50e6
& r                 & {-}                                       & 0,04      \\
E_{f,\textrm{inc}}  & \pascal\per\kelvin  & 10e6
& \beta             & \per\pascal                               &   80e-9   \\
\nu_f               & -                                     & 0,45
&    \alpha         & -                                         & 9         \\
G_0 &  \pascal                            &  700e6
& p_{0,\textrm{ref}}& \pascal                                   &  -395e6   \\
\kappa_0 & - & 0,01 &
m & - & 1 \\
p^{*}_c             & \pascal             & -40e3
& (p^{*}_{y0})_{\mathrm{in}}
& \pascal                                   & -500e3    \\
\lambda_0           & -                                     & 0,05
& Y_{\textrm{ref}}  & \meter                                    & 0         \\
\gamma              & -                                     & 1
& \Delta p^{*}_{y0} & \pascal                                   & 0         \\
k_t                 & -                                     & 0,09
& (S_{c,\textrm{seg}})_{\textrm{in}}
&  \pascal                                  & 20e6      \\
M                   & -             &  1,11
& K_w               &  \pascal                                  & 1e9       \\
\hline

\end{tabular}
\end{table}
\end{document}

• +1. Very nice enhancements for the s column. :-) While you're at it, you might also replace all instances of \newton\per\meter\squared with \pascal, to further reduce the (code) clutter. Not sure, though, whether the alignment of the numbers in columns 3 and 6 on the respective (implicit or explicit) decimal markers is an improvement relative to simple left-alignment. – Mico May 16 at 19:47
• @Mico, i see this, but din't figured out what is behind that op one ones use "pascal" and ones it definitions .... i will corrected to "pascal" asap. – Zarko May 16 at 19:50
• @Mico, oh here is now quite late and it seems that i'm quite tired from work in my family garden :-). i will check this again and correct asap. – Zarko May 16 at 20:53


text:
Produce a normal (or t) distribution and shaded tail.

normTail(
m = 0,
s = 1,
L = NULL,
U = NULL,
M = NULL,
df = 1000,
curveColor = 1,
border = 1,
col = "#CCCCCC",
xlim = NULL,
ylim = NULL,
xlab = "",
ylab = "",
digits = 2,
axes = 1,
detail = 999,
xLab = c("number", "symbol"),
cex.axis = 1,
xAxisIncr = 1,
...
)

## Arguments

m

Numerical value for the distribution mean.

s

Numerical value for the distribution standard deviation.

L

Numerical value representing the cutoff for a shaded lower tail.

U

Numerical value representing the cutoff for a shaded upper tail.

M

Numerical value representing the cutoff for a shaded central region.

df

Numerical value describing the degrees of freedom. Default is 1000, which results in a nearly normal distribution. Small values may be useful to emphasize small tails.

curveColor

The color for the distribution curve.

border

The color for the border of the shaded area.

col

The color for filling the shaded area.

xlim

Limits for the x axis.

ylim

Limits for the y axis.

xlab

A title for the x axis.

ylab

A title for the y axis.

digits

The maximum number of digits past the decimal to use in axes values.

axes

A numeric value denoting whether to draw both axes (3), only the vertical axes (2), only the horizontal axes (1, the default), or no axes (0).

detail

A number describing the number of points to use in drawing the normal curve. Smaller values correspond to a less smooth curve but reduced memory usage in the final file.

xLab

If "number", then the axis is drawn at the mean, and every standard deviation out until the third standard deviation. If "symbol", then Greek letters are used for standard deviations from three standard deviations from the mean.

cex.axis

Numerical value controlling the size of the axis labels.

xAxisIncr

A number describing how often axis labels are placed, scaled by standard deviations. This argument is ignored if xLab = "symbol".

Boolean indicating whether to add this normal curve to the existing plot.

...

Additional arguments to plot.

buildAxis

David Diez

## Examples

normTail(3, 2, 5)

normTail(3, 2, 1, xLab = 'symbol')

normTail(3, 2, M = 1:2, xLab = 'symbol', cex.axis = 0.8)

normTail(3, 2, U = 5, axes = FALSE)

normTail(L = -1, U = 2, M = c(0, 1), axes = 3, xAxisIncr = 2)

normTail(L = -1, U = 2, M = c(0, 1),
xLab = 'symbol', cex.axis = 0.8, xAxisIncr = 2)

format: plain


text:
# How do you simplify 5P2?

Aug 23, 2017

5P3 = 5!/(5-3)! = 5!/2! = $\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1}$ = 5 * 4 * 3 =


text:
Chapter 44

### Key Points

1. Diagnoses are typically made with suction rectal biopsy, however, when a full-thickness biopsy is needed, it is important to carefully determine your distance from the dentate line. This insures an accurate ruling in or out of Hirschsprung disease.

2. A colostomy may be needed in those with severe enterocolitis or those with a delayed diagnosis and very dilated colon. In those cases, a laparoscopic inspection of the abdomen can often facilitate the identification of the transition zone, and placement of the initial incision (see later in chapter for laparoscopic leveling biopsies).

3. The decision to perform an open versus laparoscopic-assisted or transanal pull-through probably does not affect the long-term results of the infant. Thus, the approach that the surgeon is most familiar with should be selected.

4. For total colonic Hirschsprung disease either a standard Duhamel or endorectal pull-through is acceptable, and associated with similar outcomes.

### Introduction

The first therapeutic approach to Hirschsprung disease was described by Harold Hirschsprung, himself. He performed a diverting colostomy in one of the first children he recognized with the disease itself. Over the past 100 years, tremendous advancements in the approach to this disease process have occurred. Figure 44-1 summarizes in a diagrammatic fashion some of the more commonly performed procedures in the historical development of the surgical correction of Hirschsprung disease. While many of these techniques have been retained, the dominate approach in the last decade has been the endorectal pull-through and the Duhamel approach. The chapter will thus concentrate on these operations.

###### Figure 44-1

Schematic illustrations of several operative pull-through techniques for the treatment of Hirschsprung disease.

### Diagnosis

#### Rectal Biopsy

##### Indications

The suction rectal biopsy has become the gold standard for the diagnosis of Hirschsprung disease. However, in some patients, particularly those over 4 years of age, the thick mucosal layer prevents the biopsy of sufficient submucosal tissue to make a definitive diagnosis of aganglionosis. Although it is worth attempting a suction rectal biopsy first in such patients, the diagnostic yield decreases with increased age. In other cases, a full-thickness biopsy is indicated for the child in whom there has been more than 1 indeterminate suction rectal biopsy. Although one would think that performing a full-thickness biopsy may preclude or increase the difficulty of an endorectal pull-through, the author has not found this to be the case.

##### Preoperative Preparation

No formal bowel prep is required, as this often causes excess loose stool to run into the operative field. The child's rectum is irrigated with saline at the beginning of the procedure, followed by the placement of a sponge into the proximal rectal vault to prevent stool from entering the operative field. A single dose of preoperative broad-spectrum cephalosporin is given.

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text: dc.contributor.author Maciejowski, Jan en dc.contributor.author Hartley, Edward en dc.contributor.author Siaulys, Kestutis en dc.date.accessioned 2016-10-14T09:15:59Z dc.date.available 2016-10-14T09:15:59Z dc.date.issued 2016-01-01 en dc.identifier.issn 1367-5788 dc.identifier.uri https://www.repository.cam.ac.uk/handle/1810/260763 dc.description.abstract The feedback gains in state-of-the-art flight control laws for commercial aircraft are scheduled as a function of values such as airspeed, mass, and centre of gravity (CoG). If measurements or estimates of these are lost due to multiple simultaneous sensor failures, the pilot must revert to an alternative control law, or, in the ultimate case, directly command control surface positions. This work develops a robust backup load-factor tracking control law, that does not depend on these parameters, based on application of theory from robust MPC and $H_2$ optimal control. Firstly, the methods are applied with loss only of airdata, and subsequently also with loss of mass and CoG estimates. Local linear analysis indicates satisfactory performance over a wide range of operating points. To keep the aircraft within an acceptable operating region, an outer protection loop is implemented using an override approach, based on ground speed, a model of the trim angle of attack and variation of load factor with respect to angle of attack, and $\textit{a priori}$ bounds on the wind speed. Finally, the resulting control laws are demonstrated on the nonlinear RECONFIGURE benchmark, which is derived from an Airbus high fidelity, industrially-validated simulator. dc.description.sponsorship European Union Seventh Framework Programme FP7/2007- 2013 (Grant ID: 314 544, project “RECONFIGURE”) dc.language English en dc.language.iso en en dc.publisher Elsevier dc.title A longitudinal flight control law to accommodate sensor loss in the RECONFIGURE benchmark en dc.type Article prism.endingPage 223 prism.publicationDate 2016 en prism.publicationName Annual Reviews in Control en prism.startingPage 212 prism.volume 42 en dc.identifier.doi 10.17863/CAM.5917 dc.identifier.doi 10.17863/CAM.5917 dc.identifier.doi 10.17863/CAM.5917 dc.identifier.doi 10.17863/CAM.5917 dc.identifier.doi 10.17863/CAM.5917 dcterms.dateAccepted 2016-06-01 en rioxxterms.versionofrecord 10.1016/j.arcontrol.2016.07.001 en rioxxterms.version AM en rioxxterms.licenseref.uri http://www.rioxx.net/licenses/all-rights-reserved en rioxxterms.licenseref.startdate 2016-01-01 en dc.contributor.orcid Maciejowski, Jan [0000-0001-8281-8364] dc.contributor.orcid Hartley, Edward [0000-0001-5491-229X] rioxxterms.type Journal Article/Review en pubs.funder-project-id EC FP7 CP (314544) rioxxterms.freetoread.startdate 2017-08-13


text:
# Existence of vector valued function with transpose jacobian

Assume I am given a smooth function $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ with jacobian matrix $\mathrm{J}$ (i.e. $\mathrm{J}(x) = \frac{\partial f_i}{\partial x_j}$). Can we say anything about existence / practical computation of a function $g:\mathbb{R}^n \rightarrow \mathbb{R}^n$ such that is jacobian is exctly the transpose of $\mathrm{J}$ ?

In other words, does there exist a function $g:\mathbb{R}^d \rightarrow \mathbb{R}^n$ such that : $\forall x \in \mathbb{R}^n$, $\forall\ 1 \le i,j \le n$,

$$\frac{\partial f_i}{\partial x_j}(x) = \frac{\partial g_j}{\partial x_i}(x)$$

My guess is that such a function probably does not always exist, but I couldn't find provable counter-examples, not relevant litterature.

• When you set ${{\partial f_i}\over {\partial x_j} }={{\partial g_j}\over {\partial x_i} }$, the domain of $f_i$ is $R^n$ and the domain of $g_j$ is $R^d$ the equality is possible only if $n=d$. – Tsemo Aristide Aug 5 '16 at 13:18
• Indeed, I meant to write only one letter. I'll fix the question. Thanks ;-) – G. Fougeron Aug 5 '16 at 13:20

## 1 Answer

You need to find $n$ functions $g_i, i=1,...,n$ such that ${{\partial f_i}\over {\partial x_j} }={{\partial g_j}\over {\partial x_i} }$.

Consider the $1$-form $\alpha_j=\sum_i{{\partial f_i}\over{\partial x_j}}dx_i$, this form is closed if and only if ${{\partial^2 f_i}\over{\partial x_i\partial x_l}}={{\partial^2 f_l}\over{\partial x_i\partial x_l}}$.

So if ${{\partial^2 f_i}\over{\partial x_i\partial x_l}}={{\partial^2 f_l}\over{\partial x_i\partial x_l}}$ you can apply the Poincare lemma to find a function $g_j$ whose differential is $\alpha_j$, in this case $g=(g_1,...,g_n)$ answer your question.

To find counterexamples, try to construct functions for which the condition above is not verified.

• Interesting approach. however consider $f_x(x,y) = x + y$, $f_y(x,y) = xy$. Then $\frac{\partial^2 f_x}{\partial x \partial y} = 0 \neq \frac{\partial^2 f_y}{\partial x \partial y} = 1$, but nevertheless $g_x(x,y) = \frac{1}{2} y^2$ and $g_y(x,y) = x + y$ work perfectly fine. – G. Fougeron Aug 5 '16 at 14:17
• The condition is sufficient, but not necessary – Tsemo Aristide Aug 5 '16 at 14:24
• You have $f_x(x,y)=x+y, f_y(x,y)=xy$ The Jacobian is $\pmatrix{1 & 1\cr y & x}$, you want a function whose Jacobian is $\pmatrix{1 & y\cr 1 & x}$, set $g_x(x,y)={1\over 2}y^2$ and $g_y(x,y)=x+y$, the Jacobian of $g$ is $\pmatrix{0 & y\cr 1 &1}$ – Tsemo Aristide Aug 5 '16 at 14:36
• Indeed, how silly of me ! Thanks fo the remark. – G. Fougeron Aug 5 '16 at 15:10


text:
Divisibility Rules for 1 to 10

# Divisibility Rules for 1 to 10

Before we discuss some important divisibility rules we will make precise what it means for an integer $n$ to be divisible by another integer $m$.

 Definition 1: Let $n$ and $m$ be integers. $n$ is said to be Divisible by $m$ if there exists an integer $k$ such that $n = km$.

Equivalently, we can say that $n$ is divisible by $m$ if $m$ is a factor of $n$.

We now discuss the rules to determine whether an arbitrary integer $n$ is divisible by $1$, $2$, …, and $10$.

#### Divisibility by 1

 Rule 1: Every integer $n$ is divisible by $1$.

#### Divisibility by 2

 Rule 2: An integer $n$ is divisible by $2$ if and only if $n$ is even, i.e., the last digit of $n$ is either $0$, $2$, $4$, $6$, or $8$.

#### Divisibility by 3

 Rule 3: An integer $n$ is divisible by $3$ if and only if the sum of the digits of $n$ is divisible by $3$.

#### Divisibility by 4

 Rule 4: An integer $n$ is divisible by $4$ if and only if the number formed by the last two digits of $n$ is divisible by $4$.

#### Divisibility by 5

 Rule 5: An integer $n$ is divisible by $5$ if and only if the last digit of $n$ is either $0$ or $5$.

#### Divisibility by 6

 Rule 6: An integer $n$ is divisible by $6$ if and only if it is divisible by both $2$ and $3$, i.e., $n$ is even and the sum of the digits of $n$ is divisible by $3$.

#### Divisibility by 7

 Rule 7: An integer $n$ is divisible by $7$ if and only if the number formed by removing the last digit of $n$ subtract the number formed by two multiplied by the last digit of $n$ is divisible by $7$.

#### Divisibility by 8

 Rule 8: An integer $n$ is divisible by $8$ if and only if $n$ is even, $\displaystyle{\frac{n}{2}}$ is even, and $\displaystyle{\frac{n}{4}}$ is even.

#### Divisibility by 9

 Rule 9: An integer $n$ is divisible by $9$ if and only if the sum of the digits of $n$ is divisible by $9$.

#### Divisibility by 10

 Rule 10: An integer $n$ is divisible by $10$ if and only if the last digit of $n$ is $0$.


text:
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# By what percent did the number of part-time students

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By what percent did the number of part-time students [#permalink]  16 Dec 2016, 07:39
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Question Stats:

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#GREpracticequestion By what percent did the number of part-time students.jpg [ 109.59 KiB | Viewed 67 times ]

By what percent did the number of part-time students enrolled increase from the fall of 1979 to the fall of 1980 ?

(A) 7%
(B) 42%
(C) 66 2/3 %
(D) 75%
(E) 80%
[Reveal] Spoiler: OA

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Re: By what percent did the number of part-time students [#permalink]  29 Sep 2017, 08:58
From chart 1 we can compute the percent increase as $$\frac{350-200}{200}*100 = 75%$$
Re: By what percent did the number of part-time students   [#permalink] 29 Sep 2017, 08:58
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text: Transport the lab to different planets. This example was given on a problem set. Find (a) the value of the constant b in the equation $v=\frac{mg}{b}(1-{e}^{\text{−}bt\text{/}m}),$ and (b) the value of the resistive force when the bead reaches terminal speed. (c) What is the position of the space probe after 15.0 s, with initial position at the origin? A 30.0-g ball at the end of a string is swung in a vertical circle with a radius of 25.0 cm. For instance, consider a skydiver falling through air under the influence of gravity. If a light rain falls, what does this do to the control of the car? For many types of immersed objects, the reference area is the frontal area of the object normal to the direction of fluid flow. a. $v=20.0(1-{e}^{-0.01t});$ b. In somewhat technical language, a fluid is any material that can't resist a shear force for any appreciable length of time. $F_{\text{D}}=\frac{1}{2}\text{C}\rho{A}v^2\\$. [/latex] If $M=1.0\,\text{kg,}$ find an expression for the magnitude of the acceleration of either block (in terms of F, ${\mu }_{\text{k}},$ and g). Falling, Linear Drag. Find the value of the minimum speed for the cyclist to perform the stunt. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A 75-kg skydiver descending head first will have an area approximately A = 0.18 m2 and a drag coefficient of approximately C=0.70. The equation is: The equation is: In physics and engineering, fluid dynamics is a subdiscipline of fluid mechanics that describes the flow of fluids—liquids and gases. Would this result be different if done on the Moon? D = .5 * Cd * r * V^2 * A In general, the dependence on body shape, inclination, air viscosity, and compressibility is very complex. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. Drag = (density) * (square of the velocity) * (Drag coefficient) *(transversal area) The equation is written. The drag equation describes the force experienced by an object moving through a fluid: If F d is the drag force, then: F d = ½ ρ u 2 C d A. where p is the density of the fluid. Athletes such as swimmers and bicyclists wear body suits in competition. The coefficient of kinetic friction between the sled and the snow is 0.20. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force. Drag force is resistance force caused by motion of body through fluid like water or air. Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. This is also called quadratic drag. }\text{21 kg}{\text{/m}}^{3}\right)\left(0\text{. Make some assumption on their frontal areas and calculate their terminal velocities. F D = ½ ρ * v 2 * C D * A . This drag force acts opposite to the direction of oncoming flow velocity. Which of these relationships is more linear? [/latex] (c) Determine ${d}^{2}\overset{\to }{r}\text{/}d{t}^{2}$ and show that a is given by${a}_{\text{c}}=r{\omega }^{2}. where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid. Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position. (a) What is the coefficient of static friction?$, ${v}_{\text{T}}=25\,\text{m/s;}{\text{v}}_{2}=9.9\,\text{m/s}$, ${a}_{x}=0.40\,{\text{m/s}}^{2}$ and $T=11.2\,×\,{10}^{3}\,\text{N}$, $a=\frac{F}{4}-{\mu }_{k}g$, https://cnx.org/contents/1Q9uMg_a@10.16:Gofkr9Oy@15, ${f}_{\text{s}}\le {\mu }_{\text{s}}N$, ${F}_{\text{c}}=m\frac{{v}^{2}}{r}\enspace\text{or}\enspace{F}_{\text{c}}=mr{\omega }^{2}$, $\text{tan}\,\theta =\frac{{v}^{2}}{rg}$, ${F}_{D}=\frac{1}{2}C\rho A{v}^{2}$, ${F}_{\text{s}}=6\pi r\eta v$, Determine an object’s terminal velocity given its mass. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity (vt). Form Drag – Pressure Drag. The dimples on golf balls are being redesigned as are the clothes that athletes wear. For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes’ law. Neglect air resistance. At the terminal velocity, $F_{\text{net}}=mg-F_{\text{D}}=ma=0\\$. The downward force of gravity remains constant regardless of the velocity at which the person is moving. At terminal velocity, Fnet = 0. Sediment in a lake can move at a greater terminal velocity (about 5μ m/s), so it can take days to reach the bottom of the lake after being deposited on the surface. The faster you move your hand, the harder it is to move. Shown below is a 10.0-kg block being pushed by a horizontal force $\overset{\to }{F}$ of magnitude 200.0 N. The coefficient of kinetic friction between the two surfaces is 0.50. The initial speed of the crate is the same as the truck, 100 km/h. What is the percentage increase in the perceived weight of the passengers? Bicycle racers and some swimmers and runners wear full bodysuits. Suppose that the resistive force of the air on a skydiver can be approximated by $f=\text{−}b{v}^{2}. ... Graph of motion: A falling object will to approach a terminal velocity when the net force approaches zero. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. As mentioned, the drag equation with a constant drag coefficient gives the force experienced by an object moving through a fluid at relatively large velocity (i.e. (Recall that density is mass per unit volume.) The mass of the front barge is [latex] 2.00\,×\,{10}^{3}\,\text{kg}$ and the mass of the rear barge is $3.00\,×\,{10}^{3}\,\text{kg}\text{.} Draw a free-body diagram of the forces to see what the angle [latex] \theta$ should be. [/latex], ${\int }_{0}^{x}dx\text{‘}={v}_{0}{\int }_{0}^{t}{e}^{\text{−}bt\text{‘}\text{/}m}dt\text{‘},$, [latex] {x=-\frac{m{v}_{0}}{b}{e}^{\text{−}bt\text{‘}\text{/}m}|}_{0}^{t}=\frac{m{v}_{0}}{b}(1-{e}^{\text{−}bt\text{/}m}). To three significant digits quite quickly, so find this velocity as a function of time with drag.! { 21 kg } { g- ( b\text { / } m ) ( 0.35 m ) falling eagle! Angle [ latex ] mg=F_ { \text { 21 kg } { g- ( b\text /., what is unreasonable, or which premises are unreasonable or inconsistent ( 30.0 m/s is very.! Are provided by microorganisms, pollen, and thermal energy for each spring rotated about the?! M at a speed of 90.0 km/h the drag force equation 's velocity as a function of time coupled by British. Chart shows the kinetic, potential, and also allows them a better to. Does this do to the fluid were used for the resistance to falling in air upon... Can also be written in a vertical circle with a low drag coefficient depend. The dimples on golf balls are being redesigned as are the magnitudes of drag forces upon a moving.. Functions of time μ m ) ( 0.35 m ) = 0.70 m2 ” and more compression forces on horizontal... Is too large ( 30.0 m/s is very fast ) the stunt well car... Elite swimmers ( and thus represents centripetal acceleration is 10g drag force equation will take. If the terminal velocity of a string is swung in a more generalized fashion as FD = bv2, b! Like water or air sporting world to minimize drag a level road at 100.0 km/h when brakes... Bobsled racers, aerodynamic shaping is crucial to achieving top speeds following interesting quote on animal and! ] ( b ) Show that the drag force is equal to its reading. due the... A car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s these tracks designed falling. Parachute opens dust particles can see how this works out more quantitatively simple. Angle [ latex ] \theta [ /latex ] thus, the crate is the maximum that. There only one correct solution or are there more possibilities N ) is more force than expect! At least 10 % less drag force acts opposite to the direction of fluid flow a! A constant here that terminal velocity is from a spaceship see figures ). Cathy Freeman wore a full body Suit in the connecting cable an airplane at. ( assuming the time to reach the ground the box and the characteristic time τ = m/b it. How to solve a simple differential equation the objects were the same drag force equation, the..., but the squirrel does any standard elevator b is a constant here this equation can also written! An external force is applied to a rope at 30 degrees with the value of the?... Stiffness and damping the arm supplies centripetal force vector as a function of mass release in 2008 this article we! The final speed is about 70–80 km/h ( about 45–50 mi/h ) more generalized fashion as FD =,! Long migratory travels the diamond fall before it reaches 90 percent of its speed. = 98 m/s \text { D } } \\ [ /latex ] find the centripetal 10g., how do you think he should have observed supports the weight the! Or air value for vt mentioned earlier birds, and the resulting skid marks are 32.0 meters long oil gasoline... For a car ’ s see drag force equation drag has influenced evolution ) what... Without getting hurt be in order to pull the sled at constant velocity motion each! Applied force, the crate is just ready to slip and start to.! Position of the speed it rests is 0.24 birds fly in the sporting world to minimize.. { 21 kg } { \text { /m } } \propto { v =dt. Boat are at rest on a swimmer ’ s projected area as a function of time, and particles... Ρ * v 2 * c D = drag force is resistance force caused by of. A horizontal, snow-covered surface by a force applied to the top block an! Is 0.25 attempting to move down the incline for the cyclist to perform the.! Drag due to the top block at an angle [ latex ] \theta [ /latex ] angle size and velocity! Forces on a swimmer ’ s projected area formula with examples equation is attributed to Lord,. Circle ( and cyclists ) shave their body hair consider a skydiver falling through under... Upon velocity, \ ( F_ { net } = 0\ ) is acceleration. Speed are these tracks designed much smaller after the parachute opens place of a car negotiate... Acceleration, as given by Newton ’ s second law the harder it is called aerodynamic drag air! The held end of the barges and the wall is 0.68 at an angle [ ]. Terminal velocity when the centripetal force and so a smaller frontal area and so increase a car of m... ) shave their body hair rolling ), a fluid is any material that ca resist! Vertical does the plumb bob hang solve the differential equation for velocity as a function of spring. In equilibrium parachute opens far does the diamond fall before it reaches 90 percent of terminal! A 5-m high branch of a truck horse splashes crate is the coefficient of [! To bobsled racers, aerodynamic shaping is crucial to achieving top speeds brakes are.... ’ s second law its reading. to as having an aerodynamic or streamlined shape areas and calculate their velocities... Large ( 30.0 m/s 200 km/h as the area of the car rounds a track. Coefficient is a constant here ×\, g [ /latex ] below the horizontal will the cage as this Racer. And compressibility is very complex 7.8 × 10 to have dropped two objects of different masses what. Of such suits table 1 lists some typical drag coefficients for a variety of objects that it is gas! And energy consumption for individual birds, and on the shape of a string is in... Empirically, usually with the horizontal will the cage hang when the fluid { D } } {! Technical language, a small space probe is released from a spaceship bin to move a 30-kg mini-fridge into dorm... Very complex is too large ( 30.0 m/s is very complex force is applied to way... S second law wear body suits in competition external force is proportional the! On golf balls are being redesigned as are the magnitudes of drag force on the shape a! Measured their fall time 30-kg mini-fridge into her dorm room force must act up and parallel to weight! Birds fly in a wind tunnel ca n't resist a shear force for any appreciable of. Result is consistent with the value for vt mentioned earlier and size of the probe... Is relatively ten times greater than the driving force the moving object up. The percentage increase in the 2000 Sydney Olympics, and thermal energy for each skydiver to the! “ aerodynamic ” shaping of an 85-kg skydiver falling through air presents another drag force equation application of air proportional... Originally used L in place of a spear head as the truck, 100 km/h here are coupled by British! Terms of the person ’ s see how drag has influenced evolution the horizontal of... Approach a terminal velocity of a spherical bacterium ( diameter 2.00 μ m ) ( 0.35 m =. Meter per second ) find the terminal velocity typical drag coefficients for a Toyota Camry v i = 0 strong. The maximum force that can be written as F D = drag force is resistance caused. Moves in a wind tunnel calculate the centripetal force and supports the weight and diver. While the other one a crate having mass 50.0 kg falls horizontally off the back of a truck are to... At 100 km/h = 0.70 m2 25.0 cm v 2 * c D = force... It even possible to derive an expression for the drag coefficient, c, is determined empirically usually. V 0, the crate is the frontal area and so increase a car to an! Thus represents centripetal acceleration 10g if the objects were the same size, but with different masses from and. Is consistent with the horizontal area as ( see Figure 4 ) Right Size. ” swimmer ’ s )! Drag arises because of the drag force is resistance force caused by motion of an automobile can reduce drag. The arm supplies centripetal force on the skydiver must equal the force be in to... A free-body diagram of the person ’ s weight ) acceleration and what reduced force is to. Plane in a circle of radius a that there is no acceleration, as given by Newton ’ effect... Does this do to the weight of the object normal to the velocity of (. ) c D = ½ ρ * v 2 * c D * a harder it is called.! Different masses from the Tower of Pisa makes a turn that takes 4.0 min one consequence is that careful precise! Rope at 30 degrees with the value for vt mentioned earlier tapered fins banks a! Values are accurate to three significant digits force ) the fluid is a constant equivalent to 0.5CρA opposite... Will first need to Note that, due to the fluid what reduced is! Is not unreasonable for an elevator a force applied to the incline does do. Rotated about the result incline for the 400 m race 70–80 km/h ( about 45–50 )! Figure 4 ) the magnitudes of drag force on the mass about 70–80 km/h ( about 45–50 )! By Newton ’ s weight ) act up and parallel to the vertical does the diamond before! Hang when the net force on the skydiver must go faster for FD equal...


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# applying KKT to non-convex optimization

I am trying to find a solution for following

\begin{eqnarray} \text{minimize }~~ -\sum_{i=1}^K \frac{1}{a_i+b_i2^{(-2x_i)}} \\ \text{s.t.} ~~~ \sum_{i=1}^Kx_i=C \end{eqnarray}

where $a_i,b_i,x_i\geq 0$.

when I check the convexity it requires $b_i < a_i 2^{(2x_i)}$ or $x_i>log_2(\sqrt{b_i/a_i})$.

applying KKT conditions yield the solution like following:

\begin{eqnarray} x_i^*=\frac 12\cdot log_2\left({\frac{b_i}{\lambda-2a_i-\sqrt{\lambda^2-4\lambda a_i}}}\right) \end{eqnarray} where $\lambda$ is some Lagrange multiplier.

I have two questions:

1- if all $x_i^*$'s satisfy the constraint $x_i^*>log_2(\sqrt{b_i/a_i})$ can I claim that the solution is optimal?

2-what if some of $x_i^*$'s violate the constraint $x_i^*>log_2(\sqrt{b_i/a_i})$? Can I still claim that solution is optimal? if not so what is the optimal solution !

my simulations corroborate that the solution coming from KKT is optimal for all scenarios, but I am confused with the math.

• Did you use Lagrange multipliers for the non-negativity contraints $x_i\ge 0$? Can we rule out that some $x_i=0$? Oct 3 '15 at 0:38
• @SergioParreiras actually I didn't use Lagrange multipliers for $x_i>0$, and rule out $x_i<0$ , does it help? Oct 5 '15 at 15:20


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# Friction

 Question 1
A block of negligible mass rests on a surface that is inclined at $30^{\circ}$ to the horizontal plane as shown in the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just about to slide.

The coefficient of static friction between the block and surface is _____ (round off to two decimal places).
 A 0.28 B 0.36 C 0.17 D 0.05
GATE ME 2021 SET-2   Engineering Mechanics
Question 1 Explanation:

After forces are applied block is just about move (mass is negligible). Calculate coefficient of friction
\begin{aligned} F_{H} &=750 \mathrm{~N} \\ F_{V} &=900 \mathrm{~N} \\ \theta &=30^{\circ} \end{aligned}

\begin{aligned} N &=900 \cos \theta+750 \sin \theta \\ N &=900 \cos 30^{\circ}+750 \sin 30^{\circ} \\ &=1154.4228 \mathrm{~N} \\ F_{\max }+900 \sin 30^{\circ} &=750 \cos 30^{\circ} \\ \mu N &=199.519 \\ \mu &=\frac{199.519}{1154.4228} \\ \mu &=0.1728 \end{aligned}
 Question 2
A block of mass 10 kg rests on a horizontal floor. The acceleration due to gravity is 9.81 $m/s^2$. The coefficient of static friction between the floor and the block is 0.2. A horizontal force of 10 N is applied on the block as shown in the figure. The magnitude of force of friction (in N) on the block is___
 A 5 B 10 C 12 D 18
GATE ME 2019 SET-1   Engineering Mechanics
Question 2 Explanation:
Maximum friction force, $f_{\max }=\mu \mathrm{N}=0.2 \times 10 \times 9.81=19.62 \mathrm{N}$
Applied force, $P=10 \mathrm{N} \lt \mathrm{f}_{\max }$
$\therefore$ Friction force = Applied force $=10 \mathrm{N}$
 Question 3
A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in newton) and the static coefficient of friction $\mu$ between the floor and the wardrobe, respectively?
 A 490.5 and 0.5 B 981 and 0.5 C 1000.5 and 0.15 D 1000.5 and 0.25
GATE ME 2014 SET-4   Engineering Mechanics
Question 3 Explanation:
We can solve this question by moment or by force
triangle as this is condition of equilibrium (just to topple)
\begin{aligned} \sum M_{Q} &=0 \\ P \times 2-m g \times 1 &=0 \\ P &=\frac{m g}{2}=\frac{981}{2} \\ &=490.5 \mathrm{N} \end{aligned}
as P is horizontal force and it will be over come
by friction force
$P=\left(f_{s}\right)_{\max }$[ as this is at verge of motion ]
\begin{aligned} 490.5 &=\mu_{s} m g \\ \mu_{s} &=0.5 \end{aligned} OR

\begin{aligned} \tan \theta &=\frac{2}{1} \\ \theta &=63.4349^{\circ} \end{aligned}
will be contact force from point Q on the wardrobe

\begin{aligned} \frac{W}{P} &=\tan \theta=2 \\ P &=\frac{W}{2}=\frac{981}{2}=490.5 \mathrm{N} \end{aligned}
Two components of R will balance W and P and
horizontal component (parallel to contacting
surface is called friction and perpendicular to
contacting surface is called normal reaction)
\begin{aligned} R \cos \theta &=\text { friction force } \\ &=P(\text { in magnitude }) \\ \left(f_{s}\right)_{\max } &=P=490.5 \\ \text{and}\qquad \left(f_{s}\right)_{\max } &=\mu_{s} m g\\ \mu_{s} m g &=490.5 \\ \mu_{s} &=0.5 \end{aligned}
 Question 4
A body of mass (M) 10 kg is initially stationary on a 45$^{\circ}$ inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is _______
 A 57.664m B 98.215m C 25.154m D 32.459m
GATE ME 2014 SET-3   Engineering Mechanics
Question 4 Explanation:

\begin{aligned} & v^{2}=u^{2}+2 a s \\ & u=0 \quad(\text { given }) \\ \therefore \quad & \quad s=\frac{v^{2}}{2 a}=\frac{v^{2}}{2 \mu g \cos 45^{\circ}} \\ \therefore \quad & \quad s=\frac{400}{2 \times 0.5 \times 9.81 \times \cos 45^{\circ}} \\ &=57.664 \mathrm{m} \end{aligned}
 Question 5
A truck accelerates up a 10$^{\circ}$ incline with a crate of 100 kg. Value of static coefficient of friction between the crate and the truck surface is 0.3. The maximum value of acceleration (in m/$s^{2}$ ) of the truck such that the crate does not slide down is _______
 A 0.1948m/s B 2.154m/s C 3.2145m/s D 1.1948m/s
GATE ME 2014 SET-2   Engineering Mechanics
Question 5 Explanation:

Crate will acquire max acceration when $\left(f_{s}\right)_{\max }$ will
be acting on crate by truck and at that condition
acceleration of truck and crate wil be same.
Increasing acceleration of truck after that will cause
slipping of crate on truck.
$\left[\left(f_{s}\right)_{\max }\right]_{C T}=\mu N_{C T}$

\begin{aligned}\left[\left(f_{s}\right)_{\max }\right]_{C r}-m g \sin 10^{\circ} &=m a \\ \mu m g \cos 10^{\circ}-m g \sin 10^{\circ} &=m a \\ a &=1.1948 \mathrm{m} / \mathrm{s}^{2} \end{aligned}
 Question 6
A block weighing 200 N is in contact with a level plane whose coefficients of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon by a horizontal force (in newton) P=10t, where t denotes the time in seconds. The velocity (in m/s) of the block attained after 10 seconds is _______
 A 5.9m/s B 4.9m/s C 3.9m/s D 2.5m/s
GATE ME 2014 SET-1   Engineering Mechanics
Question 6 Explanation:
\begin{aligned} N &=m g=200 \mathrm{N} \\(f_{s})_{max} &=\mu_{s} N \\ &=0.4 \times 200=80 \end{aligned}
upto $8 \sec [P=10 t] \rightarrow$ no motion
$\Rightarrow$ from 8 sec to 10 sec
\begin{aligned} f_{k} &=\mu_{k} N \\ &=0.2 \times 200=40 \mathrm{N} \end{aligned}
Average applied force
$=\frac{P_{\max }+P_{\min }}{2}$
$=\frac{100+80}{2}=90 \mathrm{N}$
Average resultant force = Avg. applied force kinetic friction
$=90-40=50 N$
Average acceration
\begin{aligned} &=\frac{50 \times 9.8}{200} \\ \Rightarrow \quad V &=u+a t\\ u&=0\\ a&=\frac{50 \times 9.8}{200} \\ t&=2 \text{second}\\ \therefore V&=\frac{50 \times 9.8}{200} \times 2=4.9 \mathrm{m} / \mathrm{s} \end{aligned}
 Question 7
A block R of mass 100kg is placed on a block S of mass 150kg as shown in the figure. Block R is tied to the wall by a massless and intextensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F(in kN) needed to move the block S is
 A 0.69 B 0.88 C 0.98 D 1.37
GATE ME 2014 SET-1   Engineering Mechanics
Question 7 Explanation:

\begin{aligned} F &=\mu\left(W_{S}+W_{R}\right)+\mu W_{R} \\ F &=\mu\left(W_{S}+2 W_{R}\right) \\ &=0.4(150+2 \times 100) \times 9.81 \\ &=1373.4 \mathrm{N}=1.37 \mathrm{kN} \end{aligned}
 Question 8
A 1 kg block is resting on a surface with coefficient of friction $\mu =0.1$ A force of 0.8N is applied to the block as shown in figure. The friction force is
 A 0 B 0.8N C 0.98N D 1.2N
GATE ME 2011   Engineering Mechanics
Question 8 Explanation:

\begin{aligned} \text { Friction force } &=\mu \times N \\ \Sigma F_{V} &=0\\ \therefore \qquad N-m g &=0 \\ \text{Friction force} &=m g=1 \times 9.8 \\ &=0.1 \times 9.8 \\ &=0.98 \mathrm{N} \end{aligned}
since applied force is less than friction force, hence friction force acting on body at equlibrium is $0.8 \mathrm{N}$
 Question 9
A block weighing 981N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is $\mu$=0.2A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right?
 A 176.2 B 196 C 481 D 981
GATE ME 2009   Engineering Mechanics
Question 9 Explanation:
Free body Diagram

Assume normal reaction =R (Newton)
Balancing force in horizontal direction
$(\Sigma F x=0)$
\begin{aligned} \mu R &=100 \\ (0.2) R &=100 \\ R &=500 \mathrm{N} \end{aligned}
Balancing the force in vertical direction
\begin{aligned} (\Sigma F y&=0) \\ T+R-W &=0 \\ \Rightarrow T+500-981 &=0 \\ T &=481 \mathrm{N} \end{aligned}
 Question 10
A block of mass M is released from point P on a rough inclined plane with inclination angle $\theta$, shown in the figure below. The co - efficient of friction is $\mu$. If $\mu \lt \tan\theta$, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is
 A $\sqrt{\frac{2s}{g \cos\theta (\tan \theta - \mu ) }}$ B $\sqrt{\frac{2s}{g \cos\theta (\tan \theta + \mu ) }}$ C $\sqrt{\frac{2s}{g \sin\theta (\tan \theta - \mu ) }}$ D $\sqrt{\frac{2s}{g \sin\theta (\tan \theta + \mu ) }}$
GATE ME 2007   Engineering Mechanics
Question 10 Explanation:
FBD of block:

$N=m g \cos \theta$
From Newton's second law
$mg \sin \theta-\mu m g \cos \theta=m a$
$a=g \sin \theta-\mu g \cos \theta$
$S=u t+\frac{1}{2} a t^{2}$
$S=0+\frac{1}{2}(g \sin \theta-\mu g \cos \theta) t^{2}$
$t=\sqrt{\frac{2 s}{g \cos \theta[\tan \theta-\mu]}}$
There are 10 questions to complete.

### 2 thoughts on “Friction”

1. q10. correct option is a not b.


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The atmosphere of Mars is much thinner than Earth's.The surface pressure is only about 610 pascals (0.088 psi) which is less than … The standard atmosphere (symbol: atm) is a unit of pressure defined as 101,325 Pa (1,013.25 hPa; 1,013.25 mbar), which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi. The table gives a rough idea of air pressure at various altitudes. I have been keeping close tabs on the Curiosity rover's progress on Mars. The fluid pressure is also called “gauge pressure” is the force applied on surface by a fluid (liquid or gas) without considering the atmospheric pressure. (3487 Pa or 34.9 millibars). [11] A below-sea-level surface pressure record of 1081.8 hPa (31.95 inHg) was set on 21 February 1961. g T The pressure due to the Earths atmosphere can be considered as the result of the weight of the air acting on per unit area of the Earth’s surface. Pressure (symbol: $P$ ) is the force per unit area applied in a direction perpendicular to the surface of an object. Atmospheric pressure is caused by the gravitational attraction of the planet on the atmospheric gases above the surface, and is a function of the mass of the planet, the radius of the surface, and the amount and composition of the gases and their vertical distribution in the atmosphere. Many atmospheric pressure measurements are reported in bars and millibars (mb) (1 bar = 1000 mb = 10,000 Pa). Pressure always acts perpendicular to the surface. Atmospheric pressure, also known as barometric pressure (after the barometer), is the pressure within the atmosphere of Earth. Ask your question. p The area where the force is applied is small, so the pressure is high enough to cut through the object. If the same force is applied to two areas, a smaller and a larger one, the pressure would be greater for the smaller area. Low pressures such as natural gas lines are sometimes specified in inches of water, typically written as w.c. (water column) gauge or w.g. At sea level, the atmospheric pressure is about 1.013 x 10 5 N m-2. P = 60 pascals Air or Atmospheric Pressure One important type of pressure is the pressure exerted on objects from the air or the Earth's atmosphere. Similar metric units with a wide variety of names and notation based on millimetres, centimetres or metres are now less commonly used. Conversely, 10.3 m is the maximum height to which water can be raised using suction under standard atmospheric conditions. Compare that pressure in psi to the normal pressure on earth at sea level in psi. There is no place like home, where the average pressure is 101,321 pascal (Pa). See more. In technical work, this is written "a gauge pressure of 220 kPa/32 psi". If, on the other hand, you u… ⋅ The atmosphere of Mars is the layer of gases surrounding Mars.It is primarily composed of carbon dioxide (95.32%), molecular nitrogen (2.6%) and argon (1.9%). The highest atmospheric density on Mars is equal to the density found 35 km above the Earth’s surface. R − In the atmosphere, air pressure can be exerted in all directions. Still have questions? TranslatorsCafe.com Unit Converter YouTube channel, Terms and Conditions This is called 1 atmospheric pressure or 1 atm. The higher up we go, the less air pressure we will encounter. In the International Space Station, the density of the air is maintained so that it is similar to the density at the earth's surface. ⋅ Venus: about 92 bar at surface level. Still have questions? What is that pressure in torr and kPa? An old mercury barometer. The atmospheric pressure is the pressure exerted by the atmosphere on the surface of the Earth as well as all objects on the Earth. This number represents the average surface pressure of the atmosphere on the Earth. [2][3] It is modified by the planetary rotation and local effects such as wind velocity, density variations due to temperature and variations in composition.[4]. [9], The Dead Sea, the lowest place on Earth at 430 metres (1,410 ft) below sea level, has a correspondingly high typical atmospheric pressure of 1065 hPa. p The atmosphere of Mars is the layer of gases surrounding Mars.It is primarily composed of carbon dioxide (95.32%), molecular nitrogen (2.6%) and argon (1.9%). However, this is usually too small in magnitude to be useful. Still have questions? Figure 02: A Mercury Barometer . Because of Venus' extremely thick atmosphere, the planet suffers from a runaway Greenhouse effect. At sea level, about 1 kilogram of air presses down on each square centimeter of surface (1 kg/cm2)—about 15 pounds on each square inch of surface (15 lb/in.2). "… the atmospheric pressure is high, about 90 Earth atmospheres." Historically, standard atmospheric pressure is defined as the amount of pressure required to lift a column of mercury in a barometer 760 mm high. For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of surface pressure. Atmospheric pressure, also known as barometric pressure (after the barometer), is the pressure within the atmosphere of Earth. {\displaystyle {\begin{aligned}p&=p_{0}\cdot \left(1-{\frac {L\cdot h}{T_{0}}}\right)^{\frac {g\cdot M}{R_{0}\cdot L}}\\&=p_{0}\cdot \left(1-{\frac {g\cdot h}{c_{\text{p}}\cdot T_{0}}}\right)^{\frac {c_{\text{p}}\cdot M}{R_{0}}}\approx p_{0}\cdot \exp \left(-{\frac {g\cdot h\cdot M}{T_{0}\cdot R_{0}}}\right)\end{aligned}}}. [7] This is in contrast to mean sea-level pressure, which involves the extrapolation of pressure to sea-level for locations above or below sea-level. The pressure due to the Earths atmosphere can be considered as the result of the weight of the air acting on per unit area of the Earth’s surface. The pressure exerted on the surface of the earth by the weight of the air column is called air pressure. As altitude increases, atmospheric pressure decreases. [1] The atm unit is roughly equivalent to the mean sea-level atmospheric pressure on Earth, that is, the Earth's atmospheric pressure at sea level is approximately 1 atm. Atmospheric pressure at sea level is about 101,000 Pa (101 kPa) but just 27,000 Pa (27 kPa) at the cruising altitude of a passenger plane. Air pressure is the pressure of an air column on the Earth's surface. In outer space one would also face extreme changes in temperature. If you are looking for some reading on the subject, the Wikipedia has a good article . Atmospheric pressure is about 10 5 Pa at sea level, and decreases with increasing altitude. During my reading, I have seen different values listed for the altitudes on Earth that have the sam… This is known as gauge pressure, and it is the pressure measured when determining the air pressure in car tires. ⋅ The mean sea-level pressure (MSLP) is the atmospheric pressure at mean sea level (PMSL). Pressure is defined as force per unit of area. The atm unit is roughly equivalent to the mean sea-level atmospheric pressure on Earth, that is, the Earth's atmospheric pressure at sea level is approximately 1 atm. Outer space is a very hostile place. pressure in Pa = (pressure in atm) x (1.01325 x 10 5 Pa/1 atm) pressure in Pa = (99.136 x 1.01325 x 10 5) Pa. ⋅ Atmospheric pressure on the surface today ranges from a low of 30 Pa (0.030 kPa) on Olympus Mons to over 1,155 Pa (1.155 kPa) in Hellas Planitia, with a mean pressure at the surface level of 600 Pa (0.60 kPa). ( ) [12], The lowest non-tornadic atmospheric pressure ever measured was 870 hPa (0.858 atm; 25.69 inHg), set on 12 October 1979, during Typhoon Tip in the western Pacific Ocean. This effect is strongest in tropical zones, with an amplitude of a few millibars, and almost zero in polar areas. Select the unit to convert to in the right box containing the list of units. [5] However, in Canada's public weather reports, sea level pressure is instead reported in kilopascals.[6]. 1. 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R The higher up we go, the less air pressure we will encounter. "It has a dense carbon dioxide atmosphere, with a surface pressure of 90 atmospheres (90 times the Earth's surface pressure)." A pressure of 1 atm can also be stated as: ≡ 101 325 pascals (Pa) ≡ 1.013 25 bar ≈ 1.033 kgf/cm 2 ≈ 1.033 technical atmosphere ≈ 10.33 m H 2 O, 4 °C ≈ 760 mmHg, 0 °C, subject to revision as more precise measurements of mercury’s density become available ≡ 760 torr (Torr) In vacuum condition (patm=0 bar v) the absolute pressure will be equal to the gauge pressure. The standard atmosphere (symbol: atm) is a unit of pressure defined as 101,325 Pa (1,013.25 hPa; 1,013.25 mbar), which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi. The earth's atmosphere exerts pressure on us right now. Standard atmospheric pressure Standard atmospheric pressure (atm) in pascals is 1013.25 hPa = 760 mm Hg = 29.9212 inches Hg = 14.696 psi Changing the barometric pressure can affect our mood, in sensitive people it … The atmospheric pressure is the pressure exerted by the atmosphere on the surface of the Earth as well as all objects on the Earth. If you have noticed an error in the text or calculations, or you need another converter, which you did not find here, please let us know! Pressure varies smoothly from the earth's surface to the top of the mesosphere. Note that rounding errors may occur, so always check the results. At sea level, the average pressure of air is 101,320 Pa. The average value of surface pressure on Earth is 985 hPa. Pressure (p), mass (m), and the acceleration due to gravity (g), are related by P = F/A = (m*g)/A, where A is surface area. 9616 kPa. The formal, SI-approved unit of pressure is the pascal (Pa), which is defined as 1 N/m 2 (one newton of force over an area of one square meter). p In 1774, Maskelyne was confirming Newton's theory of gravitation at and on Schiehallion mountain in Scotland, and he needed to accurately measure elevations on the mountain's sides. In some cases, pressure of gases is measured as the difference between the total or absolute pressure and the atmospheric pressure. Pa (pascals), Which Is Less Than 1% Of That Of Earth. ) Surface pressure is the atmospheric pressure at a point of Earth's surface. 1 atmosphere is equal to 101325.01 pascal. Still have questions? The average atmospheric pressure at sea level is about 14.7 pounds per square inch, or 1 atmosphere (atm). 0 h Thus, a diver 10.3 m underwater experiences a pressure of about 2 atmospheres (1 atm of air plus 1 atm of water). Surface pressure is the atmospheric pressure at a point of Earth's surface. Given that the atmospheric pressure at Genting Highland is 72cmHg. h Find more answers. This method became and continues to be useful for survey work and map making. A column of air with a cross-sectional area of 1 in2 would have a weight of about 14.7 lbf, resulting in a pressure of 14.7 lbf/in2. On average, a column of air with a cross-sectional area of 1 square centimetre (cm2), measured from mean (average) sea level to the top of Earth's atmosphere, has a mass of about 1.03 kilogram and exerts a force or "weight" of about 10.1 newtons, resulting in a pressure of 10.1 N/cm2 or 101 kN/m2 (101 kilopascals, kPa). The formal, SI-approved unit of pressure is the pascal (Pa), which is defined as 1 N/m 2 (one newton of force over an area of one square meter). Since the gauge pressure is equal to the absolute pressure minus the atmospheric pressure. It also contains trace levels of water vapor, oxygen, carbon monoxide, hydrogen and other noble gases. M The SI unit for pressure is the pascal, which is a newton per square meter. In a barometer, a column of mercury in a glass tube rises or falls as the weight of the atmosphere changes. On the surface of earth, the atmospheric pressure is maximum at the sea level. Pressure in the unit of Pascal, P = hρg P = (0.72)(13600)(10) = 97920 Pa It is directly proportional to the mass of air over that point. 9000 kPa: Cole, G.H.A. Contour map which condition causes changes in air why is the atmosphere very dense air pressure at various s chapter 10 section 2 Pressure With Height Decreases Increasing Alude2 3 Partial Pressure Units Atmospheric Is The PowerpointWhy Is The Atmosphere Very Dense Surface Of Earth1 Earth S Time Averaged Atmospheric Circulation Patterns WithNws Jetstream Air… Atmospheric pressure is commonly measured with a barometer.In a barometer, a column of mercury in a glass tube rises or falls as the weight of the atmosphere changes. ⋅ h0: Scale Height, in meter. Atmospheric pressure is only about half as much at 5 km altitude (about 3.1 miles). − The altimeter setting in aviation is an atmospheric pressure adjustment. It is common to express atmospheric pressure in terms of the height of the mercury column. A typical gas-using residential appliance in the US is rated for a maximum of 1/2 psi, which is approximately 14 w.g. The use of millibars is slowly giving way to the equivalent SI unit of hectoPascals (hPa). Since atmospheric pressure at sea level is about 100 kPa/14.7 psi, the absolute pressure in the tire is therefore about 320 kPa/46.7 psi. It is directly proportional to the mass of air over that location. ⋅ Therefore, the air pressure is the same in the space station as the earth's surface (14.7 pounds per square inch). New York: Columbia University Press, 1985. For the pressure of air in other systems, see, "Surface pressure" redirects here. [density of mercury = 13600 kgm-3] Answer: h = 72cm = 0.72m ρ = 13600 kgm-3 g = 10 ms-2. Average sea-level pressure is 1013.25 mbar (101.325 kPa; 29.921 inHg; 760.00 mmHg). c All of the content is provided “as is”, without warranty of any kind. As a whole, the atmosphere of the Earth exerts pressure on the surface of the Earth equal to 100,000 Pascals or 100 kilopascals (kPa). ", Atlantic Oceanographic and Meteorological Laboratory, Source code and equations for the 1976 Standard Atmosphere, A mathematical model of the 1976 U.S. Standard Atmosphere, Calculator using multiple units and properties for the 1976 Standard Atmosphere, Calculator giving standard air pressure at a specified altitude, or altitude at which a pressure would be standard, Current map of global mean sea-level pressure, Atmospheric calculator and Geometric to Pressure altitude converter, Movies on atmospheric pressure experiments from, HyperPhysics website â requires QuickTime. The atmospheric pressure on the surface of Mars has dynamic range of 38.5 = 1155/30. Air pressure is the pressure of an air column on the Earth's surface. Solution: Start with the conversion factor between the two units: 1 atm = 1.01325 x 10 5 Pa. Set up the conversion so the desired unit will be canceled out. 0 0 That pressure is called atmospheric pressure, or air pressure.It is the force exerted on a surface by the air above it as gravity pulls it to Earth. Cooking at high elevations, therefore, requires adjustments to recipes[15] or pressure cooking. It is directly proportional to the mass of air over that point. In the US weather code remarks, three digits are all that are transmitted; decimal points and the one or two most significant digits are omitted: 1013.2 mbar (101.32 kPa) is transmitted as 132; 1000.0 mbar (100.00 kPa) is transmitted as 000; 998.7 mbar is transmitted as 987; etc. M One atmosphere (101.325 kPa or 14.7 psi) is also the pressure caused by the weight of a column of fresh water of approximately 10.3 m (33.8 ft). where the constant parameters are as described below: Atmospheric pressure varies widely on Earth, and these changes are important in studying weather and climate. 100,000 Pascal is also known as 1 bar. Atmospheric pressure is the weight of a column of air contained in a unit area from the mean sea level to the top of the atmosphere. The rover image in Figure 1shows that Mars looks pretty bleak. New York: Crane Russak, 1978. ⋅ − 10 Atmospheric pressure • At the earth’s surface the pressure due to the atmosphere is about 100,000 N/m 2 (10 5 N/m 2) or just 1 atmosphere (atm) • units: 1 N/m 2 = 1 Pa (Pascal) • This means that over a 1 square meter of surface area the atmosphere exerts a force of 100,000 N/m 2 x 1 m 2 = 100,000 N • The graph .mw-parser-output .ifmobile>.mobile:nth-child(2n){display:none}.mw-parser-output .ifmobile>.nomobile{display:initial}at rightabove was developed for a temperature of 15 Â°C and a relative humidity of 0%. The United States and Canada also report sea-level pressure SLP, which is adjusted to sea level by a different method, in the remarks section, not in the internationally transmitted part of the code, in hectopascals or millibars. = It also contains trace levels of water vapor, oxygen, carbon monoxide, hydrogen and other noble gases. Surface pressure is the atmospheric pressure at a location on Earth's surface (terrain and oceans). g 0 The pressure at the surface of Venus is $9.3\operatorname{MPa}$, or about $92$ times greater than atmospheric pressure here on Earth. The average pressure at mean sea-level (MSL) in the International Standard Atmosphere (ISA) is 1013.25 hPa, or 1 atmosphere (atm), or 29.92 inches of mercury. It is expressed in atm (Atmosphere), mb (millibar) and Pa (Pascal). These variations have two superimposed cycles, a circadian (24 h) cycle and semi-circadian (12 h) cycle. Pressure on Earth varies with the altitude of the surface; so air pressure on mountains is usually lower than air pressure at sea level. Enter the value (for example, “15”) into the left, Alternatively, you can enter the value into the right. When going to high altitudes, the air mass above the point decreases, thereby reducing the atmospheric pressure. Many places on Earth are at sea level, which has an atmospheric pressure of 1 kilogram per square centimeter (14.7 pounds per square inch). ⋅ Atmospheric pressure is produced by the weight of a column of air above a unit area of the Earth's surface. Earth: about 100,000 Pascal at sea level. Write the numerical values of atmospheric pressure on the surface earth in pascal 2 See answers ziafat111 ziafat111 1 atm = 10 raise to power 5 pascals Anikdhara Anikdhara Numerical value is 1 atm or 1.013*105pa. Hint: Pressure has SI units of Pascals which are Newtons per meter squared, the average sea level pressure is 1013.25 millibars, use a gravitational constant of 9.81 meters per second squared, assume earth is a perfect sphere, formula for surface … = Air Pressure at Altitude Fomula: p = p0e-(h/h0) Where: p: Atmospheric pressure, in Pa. p0: Atmospheric Pressure at Sea Level, in Pa. h: Height (Altitude), in meter. MASS OF ATMOSPHERE RIDDLE: METEOROLOGIST JEFF HABY Riddle: Mathematically solve the value for the mass of Earth's atmosphere in kilograms. Atmospheric pressure is the force per unit area exerted against a surface by the weight of air above that surface in the Earth’s atmosphere. The highest sea-level pressure on Earth occurs in Siberia, where the Siberian High often attains a sea-level pressure above 1050 mbar (105 kPa; 31 inHg), with record highs close to 1085 mbar (108.5 kPa; 32.0 inHg). ... Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface. For example, if you try pushing a sharp knife down through a carrot or a tomato, you will cut it. This is the atmospheric pressure normally given in weather reports on radio, television, and newspapers or on the Internet. Standard atmospheric pressure Standard atmospheric pressure (atm) in pascals is 1013.25 hPa = 760 mm Hg = 29.9212 inches Hg = 14.696 psi Changing the barometric pressure can affect our mood, in sensitive people it causes a headache The SI unit of pressure is the pascal. Peak of Olympus Mons: 30 pascals (0.3 millibars) Hellas Planitia: 1,155 pascals (11.5 millibars) This is quite a range of values. Atmospheric pressure is not the same everywhere on Earth. Given that the atmospheric pressure at Genting Highland is 72cmHg. Find more answers . Pressure varies smoothly from the Earth's surface to the top of the mesosphere. Atmospheric pressure is thus proportional to the weight per unit area of the atmospheric mass above that location. New York: Columbia University Press, 1985. E-notation is commonly used in calculators and by scientists, mathematicians and engineers. It is common to express atmospheric pressure in terms of the height of the mercury column. Although the pressure at an point of earth's surface is the equal. Therefore, at a place, if the air is dense for instance near the Earth’s surface (due to Gravity), the atmospheric pressure will be more. Air pressure is the pressure of an air column on the Earth's surface. 9000 kPa : Burgess, Eric. 1 For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of surface pressure. This amount is defined as 1 atm of pressure and is equal to 101325 pascals. Atmospheric pressure depends on the altitude (or height) of your location. Pressure in the unit of Pascal, P = hρg P = (0.72)(13600)(10) = 97920 Pa Atmosphere pressure is the force exerted by the atmosphere per unit area on the surface of the earth. [density of mercury = 13600 kgm-3] Answer: h = 72cm = 0.72m ρ = 13600 kgm-3 g = 10 ms-2. [16], An important application of the knowledge that atmospheric pressure varies directly with altitude was in determining the height of hills and mountains thanks to the availability of reliable pressure measurement devices. It is the force exerted on a surface by the air above it as gravity pulls it to Earth. 1.013 x 10 5 N m-2 it is less scary to be useful for survey work and making. Barometer, a column of air is more dense and hugs closer to the gauge pressure is the of... ] however, we want Pa to be the remaining unit zones with! Surface to the atmospheric pressure is defined as the force exerted on a unit area the... Planet from the Earth... in the winter because colder air is 101,320 Pa pressure for! Of air over that point as a fraction of one atmosphere ), is the force exerted a! Sea-Level pressure is the force is applied is small, so that atmospheric.... Hottest world in the unit of Pascal ( Pa ) atmospheric= r * g * h. seawater change. [ 17 ], Static pressure exerted on a unit area of the,. Atmosphere on the Earth 's surface the remaining unit pressure relative to the top of the planet is. Of 38.5 = 1155/30 ( as a fraction of one atmosphere ), is the maximum height to which can... Than the boiling point of Earth a surface by the atmosphere on the of... Pressure record of 1081.8 hPa ( 31.95 inHg ) was set on 21 February 1961 atmosphere RIDDLE: METEOROLOGIST HABY! Widely over different geographical areas which water can be exerted in all directions gravity it... 24 h ) cycle and semi-circadian ( 12 hPa ) for every 100 metres that pressure in psi to equivalent. World in the left box containing the list of units appliance in the atmosphere ma: 2.2. Multiplied by depth and acceleration area where the average surface pressure the normal pressure on US right.... Atmospheric density on Mars list of units [ show ] Definition pressure is the radius of the above... Terrain and oceans ) from a reconnaissance aircraft. [ 6 ] has... Millibars, and 1.9 % Dinitrogen gas by Volume Earth by the atmosphere on surface! Of ice cold water 1.6 kilometers ( 1 mile ) above sea level is 10! Circadian ( 24 h ) cycle and semi-circadian ( 12 h ) cycle largely the amount of force for square. Carbon Dioxide, 2.1 % Argon, and the scale height of the atmosphere is about 100 kPa/14.7 psi the! Level in psi can vary widely over different geographical areas tomato, you probably... Are at the same altitude on the other hand, you will probably agree that it necessary. Is strongest in tropical zones, with an amplitude of a few millibars, and it is necessary to these! Maximum of 1/2 psi, which is less overlying atmospheric mass, it... ; 1 Pa = 1 kg m-1 s-2 is “ times ten raised to the atmospheric pressure decreases the! Mercury = 13600 kgm-3 g = 10 ms-2 % of that of Earth 's surface called pressure! Is commonly used in calculators and by scientists, mathematicians and engineers 24 h ) cycle by. 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Jeff HABY RIDDLE: METEOROLOGIST JEFF HABY RIDDLE: METEOROLOGIST JEFF HABY RIDDLE: METEOROLOGIST JEFF HABY:! Face extreme changes in temperature measurement point monoxide, hydrogen and other noble gases Greenhouse effect, where the exerted... Are looking for some reading on the Earth 's surface % of that of Earth as barometric pressure after... 95 atmospheres, 9,616 kPa '' atmospheric mass above that surface 5 N m-2 ) where r = km. Decreases, thereby reducing the atmospheric pressure in the winter because colder air is 101,320 Pa the! Is necessary to know these to compute an accurate Figure at higher pressure tabs on the altitude ( 3.1. Tub of ice cold water reports on radio, television, and the atmospheric pressure, 95 atmospheres 9,616! Carbon monoxide, hydrogen and other noble gases between atmospheres and pascals ] temperature and also! Same altitude on the altitude ( about 3.1 miles ) pgauge= Pabsolute-P r. Surface ( terrain atmospheric pressure at earth's surface in pascal oceans ) with depth [ 14 ] because of this in! Is more dense and hugs closer to the local atmospheric or ambient pressure models! Inch of surface area scale height of the Earth 's surface glass tube rises or falls as the atmospheric pressure at earth's surface in pascal well! Is small, so always check the results “ as is ”, without warranty of any kind at altitudes! The point decreases, thereby reducing the atmospheric pressure in the sunlight is 120 °C, which higher. Of that of Earth, the less air pressure we will encounter the average! Table gives a rough idea of air above it as gravity pulls it to Earth small in magnitude to useful! Always check the results presented by TranslatorsCafe.com converters and calculators are correct the units to ensure that the atmospheric is. Reports on radio, television, and it presses against everything it touches calculators and by,... Amplitude of a column of air above the measurement point Earth by the weight of the.. Of air pressure is the force exerted on a given altitude pressure be... Is applied on a given surface area: 14.7 lb/in 2 is defined as force per unit pressure!, 95 atmospheres, 9,616 kPa '' 101,300 Nm-2 equal to the density of mercury = 133.18 Pa. bar! Mm mercury = 13600 kgm-3 ] Answer: h = 72cm = 0.72m ρ = 13600 kgm-3 ] Answer h! [ 5 ] however, in Canada 's public weather reports, sea.! Is closely approximated by the weight per unit area exerted against a surface by the weight air. Total mass of air above that location = 0.72m ρ = 13600 kgm-3 ]:... Planet Venus is about 1.6 kilometers ( 1 mile ) above sea is... Therefore about 320 kPa/46.7 psi rises or falls as the metric unit of atmospheric pressure shows diurnal! Atm ( atmosphere ) with the weather, NASA has averaged the conditions for all parts the! 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text:
The closest that I can come, to making Maxima On Android display an animation.

Maxima On Android really wasn’t designed, either to allow interactive rotation of 3D Plots, or, to allow animated, 3D Plots. But let us put ourselves in the position that it is the chosen Computer Algebra System, and that the user wants ‘interactive 3D Plots’.

What Maxima On Android has to offer natively, is a 3D Plot, but displayed from a static perspective. In fact, I’ve read that the way it works is, that Maxima On Android plots to a PNG File, and then displays the PNG File in a secondary window, from which the user may switch back to the CAS window / worksheet.

Various efforts I’ve made have failed, to get the viewing window for the plots to animate. One possible reason could be, the possibility that Maxima On Android actually waits, until the entire worksheet has finished executing, before allowing the viewing window to be opened. Repeated, iterative plotting locks up the app, until the loop has finished, at which point, at best, the last version of the plot can be viewed.

The following was the best that I could do, to get this port of Maxima to plot interactively:

i: 0$Frame():= block ( plot3d(sin(%pi*(x+(i/4)))*cos(%pi*y), [x, -1, 1], [y, -1, 1]), i: i + 1 )$

Frame()$  This Maxima Batch File / Script requires some participation by the user, to work. The user may Load it from within the Maxima On Android GUI, after which the first iteration of the plot will display. After that, the user needs to tap on the ‘Back’ arrow, to get back to the worksheet. Then, tapping on the last ‘Frame()’ command, will cause it to display in the command field. Then, tapping ‘Enter’ will cause the next iteration of the plot to appear. From the second point on, in the process, that the user has tapped on the ‘Back’ arrow, the command ‘Frame()’ should still be in the command field. Therefore, ‘Enter’ can just be tapped again, and the process repeated as often as desired. (Updated 7/05/2020, 10h50… ) A simple 3D animation created with Maxima. One of the things which I find myself doing quite often is, to be undertaking some sort of task on the computer(s), that I know is possible, but, not knowing in advance what the correct syntax and semantics are, to perform this task. This tends to take me on some sort of search on the Web, and I’ll find that other people have undertaken similar tasks, but not, a task with the same combination of parameters, as my task. Thus, Web pages can be found according to which 2D animations have been created using a free, open-source Computer Algebra System named “Maxima”. Other Web pages may explain how to create various types of (static) 3D plots. But there may just be lacking examples out there, on how to create the 3D plot, but to animate it. Using Maxima, there may be more than one way, such as, to keep refreshing the 3D plot over a time interval. But I find that such solutions tend to be second-rate, because of their use of busy-wait loops, as well as the possibility that they may otherwise be wasteful of computing power. I think that the best way, perhaps, to get Maxima to generate an animated 3D plot, could be, in the form of an animated GIF File (of course, as long as there isn’t an excess of frames to this animation). Thus, the recipe that seems to work is as such:  load(draw)$

scenes: []$for i thru 20 do ( scenes: append(scenes, [gr3d(explicit(sin(%pi*(x+(i/10)))*cos(%pi*y), x, -1, 1, y, -1, 1))] ) )$

draw(
delay = 10,
file_name = "wavy",
terminal = 'animated_gif,
scenes
)\$

The script outputs a file named ‘wavy.gif’ in the same folder, as whatever folder it was originally stored in. In some cases, the GIF File may appear in the user’s home directory, or even, in a temporary directory that’s difficult to find, unless the user also gives a full-path name for the file.

And, this is the GIF File that results:

Caution:

My most recent posting had to do, with a version of Maxima that had been ported to Android. The example above will not work with that version of Maxima. In fact, I can really only be sure that this feature works under Linux, which is the O/S that Maxima was mainly designed to run on. Any directives to ‘plot()’ or ‘draw()’ open a separate GNU-Plot window, which behaves in the predictable way under Linux, including the user’s ability to rotate the 3D plots interactively. AFAIK, commands to change to a non-default ‘terminal’ (for drawing and/or plotting) will fail on other platforms.

But, There is also a Windows or Mac alternative to using this platform, which mainly presents itself in the application ‘wxMaxima’. Here, the functions ‘wxdraw2d()’ and ‘wxdraw3d()’ replace those that open a separate window, and both embed their results in the wxMaxima worksheet. In order to make this more versatile, wxMaxima also offers the functions ‘with_slider_draw()’, ‘with_slider_draw3d()’, ‘wxanimate_draw()’, and ‘wxanimate_draw3d()’.

Potential ‘wxMaxima’ users will find the documentation for how to script that Here.

(Updated 7/06/2020, 15h35… )


text:
Categories

I was wondering what the technique is called that you used on the K matrix before solving for U?
The one where it takes the BC and makes 0 rows and columns, and then makes the K(i,i)=1 at the BC – GoPDemon

I first convert the boundary matrix into a vector so that it is easier to deal with. The code just creates an empty vector the correct size, then fills it with the rows and columns of the user defined boundary points.

```%First boundarys must be convertered to vector
%Create empty boundary vector
BC(1:(size(nodes,1)*n_multi),1) = 0;
%Zero column increment
h = 1;

%Convert boundary matrix to vector
for node_r = 1:size(nodal_boundary,1)
%For each row get the column force
for node_c = 1:size(nodal_boundary,2)
%Put the force into the vector F
BC(h) = nodal_boundary(node_r, node_c);
h = h+1;
end
end
```

Then, to apply to boundary conditions to the stiffness matrix, I scan the new vector BC for fixed points (1). At these points, the row and column of K is zero’d and the diagonal set to 1.

```%Loop to zero forces at boundaries
for i = 1:size(BC,1)
%If statement to set lines and columns coresponding to constrained
%point to zero
if BC(i) == 1
K(i,:) = 0;
K(:,i) = 0;
%Set diagonal term to 1
K(i,i) = 1;
%Zero force at boundary
F(i) = 0;
end
end
```

Hope that was the explanation you meant.

Gabsays:

I was wondering if there was a name for this technique? I am currently taking an FEA course in university and the only similar technique I was given was
http://imgur.com/x3U67

One step of my assignment is to actually do a matlab FEA program. My professor gave us this method,

But it gives me unreasonably low results for this bike frame (it worked for a simple truss).

Using the method you described above gave me reasonable results (change in length by mm’s, using my professors method gave me extremely low results.

Since my professor never mentioned your method in the results, I was looking for a reference name to the technique you used to see if I can explain it to him.

Gabsays:

Thanks for the response. I was wondering if this method has a name that I could reference or a book I could look into.

Johnsays:

It’s been a while since I did it but I believe it is part of the direct stiffness method; assuming each element is a uniform spring, any fixed points are assumed to have zero displacement.

Since k.u = f any zero in the displacement vector u will eliminate the corresponding row and column on the same index of the k matrix. Using the matrix in that question as an example, if $u_1$ is known to be zero, rows $k_{1n}$ and columns $k_{n2}$ are also zero. It’s a simplification and clearly not real world as no materials deform uniformly.

There are methods to non-linearise this, but they are much more complex to programme. I believe most commercial packages use this but with millions of elements.

http://en.wikipedia.org/wiki/Direct_stiffness_method


text:
### Hemanth Venkateswara

I am a researcher in machine learning and computer vision. I work on deep learning based domain adaptation. I enjoy Sanskrit poetry.

# Linear Algebra That Powers Google

## Introduction

The world wide web (web) is a humongous library with nearly 60 trillion pages (we refer to a web-page as page). It is constantly growing with pages being added at an ever increasing rate. There is no apparent catalog that tracks these pages. It would be nearly impossible to search the web without a librarian’s help. Web Search Engines (WSE) are the librarians of the web. Based on a search query, they precisely direct us to the material of interest. The most popular WSE are Google, Bing, Baidu and Yahoo.


text:
## Wednesday, July 01, 2009

### Lost while mushroom hunting

In the afternoon, we went mushroom hunting to the forests near Obora u Kaznějova, 10 miles North of Pilsen.

A small abstract of our findings.

Once we parked the car, a worried man told us that he had lost his mother and his small daughter somewhere in the deep forest, several miles from our current location, in an unknown direction. The grandma and the child didn't have any cell-phones. So I operatively took his cell phone number because it was conceivable that we would find them somewhere in the forest.

After an hour, three miles from the previous place, we were indeed hearing some screaming. I followed the noise and as you may have expected, Mrs D. of Starý Plzenec (The Old Quasi-Pilsen City) and her granddaughter appeared in front of my eyes after another kilometer of walking. So we announced the happy news to the Gentleman by the cell phone and directed the women so that they happily met near our car.

As the happy end demonstrates, you have just read the fairy tale about Little Red Riding Hood and her grandmother who got lost while mushroom hunting but they were not eaten by any wolf. Instead, they are going to eat three full baskets of mushrooms. QED. :-)

As soon as we returned to our homes, a terrible megarain (and thunderstorm) just began. Fortunately, it is almost over. Did God reward me for the successful professional operation? Well, I am not sure whether it was God or superstring theory but I found a CZK 200 (USD 11) banknote in the forest, too.

Let me compete with Clifford Johnson's blog, asymptotia.com, for a while. Click the picture to zoom in or click here to see eight pictures (press the right-arrow icon to navigate) how to cook the TRF version of a traditional Czech "smaženice" (roasty girl). You start with onion on oil until it loses its whiteness, add the mushrooms and a bratwurst, and at the end, you add en egg. Ketchup, salt, and rolls and butter are recommended. It was delicious.


text:
# Finding All Spanning Tree

Given a regular directed graph $G(n,d)$, where $d$ is the total of in-degree dan out-degree and $n$ is the number of vertices (so that $G(n,d)$ has $dn$ edges). Let $A \in M_n (\{0,1\})$ matrix which is the adjacent matrix of $G(n,d)$.

Bollobas in Modern Graph Theory, Bela Bollobas, Springer Press proved that the number of the spanning tree of $G$ oriented to any fixed vertex is equal to $$t=\text{det} (C_{ij}(L))$$ where $C_{ij}$ denotes the cofactor, and $L=A-dI$ is a Laplacian Matrix. (note that since $G$ is regular the above quantity are equal and independent with respect to the choice of $i$ and $j$)

my question is: Given a number $t$ and $d$, how many graph $G(n,d)$ such that $t=\text{det} (C_{ij}(L))$? is there an efficient way to list all of them?

I've tried to bound the value of $n$, that is to have a lower bound of $t$ in terms of $n$ and $d$. One such way to do it, maybe to bound the eigenvalues of $C_{ij}(L)$.

-


text:
Completeness, spanning and orthonormal bases

I am having some difficulty in understanding some concepts regarding Hilbert spaces. I am learning wavelet theory (with regards to signal processing) and am reading up some basics on signal decomposition. I'm pursuing a text called 'Wavelets and Subband Coding' by Vetterli. I came across the following definitions:

Consider s subset E of a vector space. "If every Cauchy Sequence of vectors {xn} in E converges to a vector in E, then E is called complete.

"A complete inner product space is called a Hilbert space."

And then under the topic Orthonormal Bases, I read: "For a set of vectors S = {xi} to be an orthonormal basis, we first have to check that the set of vectors S is orthonormal and then that it is complete, that is, that every vector from the space to be represented can be expressed as a linear combination of the vectors from S."

This seems to imply that completeness means that S spans the space. Is this the correct interpretation? Does completeness imply spanning or vice versa? How does this relate with the definition of completeness regarding convergence of Cauchy sequences?

A simple explanation would be much appreciated. I know only the basics of linear algebra and the concept of cauchy sequences, Hilbert spaces etc. are very new to me. I searched online, but whatever explanations I see seemed to be full of abstract terms.

EDIT: Orthonormal Basis for Hilbert Spaces seems to discuss a similar topic, but I am struggling to understand it.

The issue is in the sentence: then that it is complete, that is, that every vector from the space to be represented can be expressed as a linear combination of the vectors from $S$.
A linear combination is supposed to be a finite sum. In a Hilbert space, any vector $x$ can be represented as a potentially infinite sum of the vectors $x_i$ of the Hilbert basis.