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# two springs have their respective force constants k1 and k2

## two springs have their respective force constants k1 and

two springs have their respective force constants k1 and k2 both are streached till their elastic potential energies are equalif the stretching forces 5ccwhwee -Physics - TopperLearning.com

## Two springs have their respective force constants K1 and

Jun 06, 2019  Two springs have their respective force constants K1 and K2. Both are stretched till their elastic potential energies are equal. If the stretching forces are F1 and F2 find F1/ F2

## Two springs of spring constants k1 and k2 are joined in

First off, for this type of question, you need to remember the fact that the force from a spring comes from how long it is stretched multiplied by its constant, k1 or k2. In series, both of these springs would not stretch equally, as they have different spring constants. So, we can come up with 2 equations f1= k1*x1-(1) and f2=k2*x2-(2). Assuming a force F is applied on this system, the springs will stretch a combined

## Two springs have their force constant as k1 and k2 (k1 > k2).

TCYonline Question & Answers: get answer of Two springs have their force constant as k1 and k2 (k1 > k2). When they are stretched by the same force, For full functionality of this site it

## Two springs and force questions? - Yahoo Answers

May 06, 2012  Two springs, with force constants k1 and k2 are connected in series, as shown in the figure. How much work is required to stretch this system a distance x from the equilibrium position? (Use any variable or symbol stated above as show more Two springs, with force constants k1 and k2 are connected in series, as shown in the figure.

## Solved: Two Springs Have Spring Constants K1 And K2 > K1

Question: Two springs have spring constants k1 and k2 > k1. Connected as shown, they act like one spring Two springs have spring constants k1 and k2 > k1. Connected as shown, they act like one spring. A) Compute the spring constant of the combination. Express the answer in terms of the variables k1 and k2.

## Equivalent spring constant . - Physics Forums

Feb 27, 2011  Equivalent spring constant.. Homework Statement This is a general doubt i have..when two springs of spring constants k1 and k2 are connected in series the equivalent k is given by 1/k=1/k1+1/k2 and when they are in parallel it is k=k1+k2..Now in these cases do we assume that the natural length of the springs are same? Homework Equations

## Two springs have force constants k1 and k2(k1 k2) On which

Two springs have force constants k1 and k2(k1 k2) On which spring is more work done,if(i)they are stretched by the same force (ii) they are stretched by the same amount - Physics -

## two springs have force constants k1 and k2 where k1 k2 on

two springs have force constants k1 and k2 where k1 k2 on which spring more work is done if they are stretched by the same magnitude of force - Physics - TopperLearning.com

## Two springs, with force constants k1=175N/m and k2=270N/m

Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x? asked by ana on March 27, 2017

## Two springs have their force constant as k1 and k2 (k1 > k2).

TCYonline Question & Answers: get answer of Two springs have their force constant as k1 and k2 (k1 > k2). When they are stretched by the same force, For full functionality of this site it

## Four massless springs whose force constants are 2k, 2k, k

Two springs of force constants k and 2k are connected to a mass as shown in figure. a body having mass m is attached with two springs having spring constants k1 and k2. asked Apr 25 in Physics by Ankitk (73.9k points) A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students

## Two springs and force questions? - Yahoo Answers

May 06, 2012  Two springs, with force constants k1 and k2 are connected in series, as shown in the figure. How much work is required to stretch this system a distance x from the equilibrium position? (Use any variable or symbol stated above as show more Two springs, with force constants k1 and k2 are connected in series, as shown in the figure.

## Solved: Two Springs Have Spring Constants K1 And K2 > K1

Question: Two springs have spring constants k1 and k2 > k1. Connected as shown, they act like one spring Two springs have spring constants k1 and k2 > k1. Connected as shown, they act like one spring. A) Compute the spring constant of the combination. Express the

## A 3.00-kg block is connected to two ideal horizontal

Jan 08, 2017  A 3.00-kg block is connected to two ideal horizontal springs having force constants k1 = 25.0 N/cm and k2 = 20.0 N/cm The system is initially in equilibrium on a horizontal, frictionless surface.

## Finding the period with mass attached to two springs

Apr 15, 2010  When the two springs of spring constant k1 and k2 are connected in series, they behave like a single spring of equivalent spring constant k = k1*k2/(k1+k2).

## If two springs have their spring constants as k1 and k2

Answer to: If two springs have their spring constants as k1 and k2 (k1>k2 . When they are stretched by same force what will be the relation between

## Two springs have force constants k1 and k2(k1 k2) On which

Two springs have force constants k1 and k2(k1 k2) On which spring is more work done,if(i)they are stretched by the same force (ii) they are stretched by the same amount - Physics -

## Two springs, with force constants k1=175N/m and k2=270N/m

Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x? asked by ana on March 27, 2017

## Two springs, of force constants k1 and k2 are connected to

Mar 23, 2018  Two springs, of force constants k 1 and k 2 are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k 1 and k 2 are made four times their original values, the frequency of oscillation becomes (a) 2f (b) f/2 (c) f/4 (d) 4f.

## Force constants k1 and k2 of two springs are in the ratio

Force constants k1 and k2 of two springs are in the ratio 5:4 .They are stretched by same length .If potential energy stored in one spring is 25 J then potential energy stored in second spring is

## Two springs have 🚩 force constants k1 and k2. on spring

Correct answer to the question: Two springs have force constants k1 and k2. on spring work done is more if 1) are streched by the same force and - brainsanswers.in

## Two springs, with force constants k1=170N/m and k2=255N/m

Sep 24, 2019  Two springs, with force constants k1=170N/m and k2=255N/m, are connected in series, as shown in (Figure 1).? When a mass m=0.75kg is attached to the springs, what is the amount of stretch, x? Express your answer to two significant figures and include appropriate units.

## Two springs of constants k1 and k2 are connected in series

Dec 19, 2011  Two springs of constants k1 and k2 are connected in series to a block a that vibrates in simple harmonic motion with a period of 5 s. When the same

## two springs with constants k1 and k2 are in a series Using

Two springs with constants k1 and k2 are in a series Using the fact that force balance must apply everywhere, what is the equivalent spring constant of a single spring? That is, if you had to swap out a single spring for the two springs that are in series, what would its spring constant need to be to generate the same force at any displacement?

## What is the k constant of a system of two springs hung

Nov 06, 2011  Here's the trick: when you hang springs in a series like this, they have the same force applied across each spring, so F1 = F2 = F. F1*k1 + F2*k2 = F*k1 + F*k2 = F*(k1 + k2) F*(k1 + k2) = Dtotal. F*Ktotal = Dtotal. Ktotal = k1 + k2. So Ktotal is just the sum of k1 and k2.

## Two springs have force constant K1&K2 (K1 greater than K2

Answer to: Two springs have force constant K1&K2 (K1 greater than K2) . Each spring is extended by the same force F. If their elastic potential

## Two weightless springs with force constants k1 and k2 are

Feb 04, 2018  Mass * Acceleration due to gravity = (k1+k2) * dx/dt. Where x is the distance the mass is moved to due to the weight of the mass m ; ) →m*9.81 * dt = (k1+k2) dx. Integrating On both sides and assuming the initial conditions to be zero, →m*9.81*t =

Nov 30, 2013  1) A 3.00-kg block is connected to two ideal horizontal springs having force constants k1 = 25.0 N/cm and k2 = 20.0 N/cm (the block is at the middle of two springs). The system is initially in

## Two springs with spring constants K1=1500 N/m and K2=3000

Two springs with spring constants K1=1500 N/m and K2=3000 N/m are stretched by the same force. Two springs with spring constants K1=1500 N/m and K2=3000 N/m are stretched by the same force. The ratio of potential energy stored in spring will be 18000+ students are using NEETLab to improve their score. What about you? Solve Previous Year

## Two springs of force constants k1, and k2, are connected in

Two springs of force constants k1, and k2, are connected in series. The spring constant of the combination is (A) k1+k2 (B) (k1+k2/2) (C) (k1+k2/k1 k2) (D) (k1k2/k1+ k2). Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and

## Two springs, of force constants k1 and k2, are connected

Two springs, of force constants k1 and k2, are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes from Physics Oscillations Class 11 Mizoram Board

## What is equivalent spring constant - Answers

Sep 13, 2011  For two springs in parallel the net force is equal to the sum of the force of each spring. Let one spring have spring constant k1 and the second k2 . Feq = keqx = (k1+k2)x keq = k1+k2 For two

## two springs have force constant K1 and K2 (k1k2) EACH

two springs have force constant K1 and K2 (k1k2) EACH SPRING IS EXTENDED BY SAME FORCE f , IF THEIR elastic potential energy are E1 and E2 then, E1/E2 IS = - Physics -

## Two springs X and Y have spring constants k and 2k

They have spring constants 2k and k respectively.The springs are stretched, separately, by a force that is gradually increased from zero up to a certain maximum value, the same for each . asked by Taylor on October 28, 2014; physics. A single mass (m1 = 3.1 kg) hangs from a spring in a motionless elevator. The spring constant is k = 338 N/m.

## Work Power Energy - Sakshi Education

Two springs have their force constants k1 and k2 (K2>K1). When they are stretched by the same force, work done is 1) Same in both the springs 2) More in spring K1 3) More in spring K2 4) None 4. A lorry and a car moving with same KE are brought to rest by applying the same retarding force.

## Physics Conceptual Questions Flashcards - Quizlet

two springs are identical except for their spring constants k2 > k1. If the same force is used to stretch the springs,

## 2 springs,with force constants k1=125N/m and k2=275N/m are

Mar 29, 2017  2 springs,with force constants k1=125N/m and k2=275N/m are connected.Mass=0.95kg is attached to the springs,what is the amount of stretch,x?

## Calculate Rate of Springs in Parallel and in Series

You have four springs that have a 20 lbf/in (pounds of force per inch) rate. To calculate the spring rate in parallel you must add up the spring rates of all the springs as shown in the formula below. Keq = k1 + k2 + k3 + k4. Keq = Equivalent Rate; k1 = Rate 1; k2 = Rate 2;

## Two springs which have force constants 1.16 N/m and 2.99 N

Oct 19, 2012  For springs in series, the Force is the same but the X is additive. K1 =1.16 Nmm K2 = 2.99 N/mm Ke =Equivalent. Stiffness. X = X1 + X2 Ke = F/X X= F/Ke X1 = F/K1 X2 = F/ K2. F/Ke = F/K1 + F/K2. 1/Ke = 1/K1 + 1/ K2. Ke = K1 K2 /(k1 + K2) = (1.16x2.99)/(1.16 + 2.99) =

## Two springs have spring constants k1 and k2 > k1

Two springs have spring constants k1 and k2 > k1. Connected as shown in (Figure 1) , they act like one spring. 1.) Compute the spring constant of the combination.

## Mechanics - 159881 - two springs have force constants k 1

two springs have force constants k 1 and k 2. Both are stretched till their elastic energies are equal .If the stretching forces are F 1 and F 2 then F 1 : F 2 is See Attachment

## Consider two massless springs connected in parallel

Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thin, vertical rod. A constant force of magnitude F is being exerted on the rod. The rod remains perpendicular to the direction of the applied force so that the springs are extended by the same amount.

## JEE Main & Advanced Physics Simple Harmonic Motion

question_answer1) Two bodies M and N of equal masses are suspended from two separate massless springs of force constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude M to that of N is [IIT-JEE 1988; MP PET 1997, 2001; MP PMT 1997; BHU 1998; Pb.

## Two springs of force constants K1 and K2 (K1> K2) are

Solution Two springs of force constants K1 and K2 (K1> K2) are stretched by same force. If W1 and W2 be the work done stretching the springs then Concept: Elastic Energy.

## Two springs of constants k1 and k2 have equal highest

Two springs of constants k1 and k2 have equal highest velocities - 12204442 p 30 periodic table of elements and report Two solid spherical planets of equal radii R having masses 4 M and 9 M their centre are separated by a distance 6 R. A projectile of mass m is sent fr mass 4 M towards the havier planet. what is the distance r of the

## What are the 2 forces involved in a stretched spring - Answers

For two springs in parallel the net force is equal to the sum of the force of each spring. Let one spring have spring constant k1 and the second k2 . Feq = keqx = (k1+k2)x keq = k1+k2 For two

## If 2 springs of spring constants k1 k2 are connected in

(.550)(9.8)/(. 063)=85.6 N/m Explain why kx=mg F = -mg = - kx, where x is the displacement from the natural equilibrium length of the vertical spring Explain why the k value for the two springs in parallel system is equals to k1+k2. With two springs in parallel, pushing down with the same displacement gives you twice the force pushing back.

## Two light springs of force constants k1 and k2 each 3.2 Nm

Two light springs of force constants k1 and k2 each 3.2 Nm-1 and a block of mass 200g are in one line AB on a smooth horizontal table such that one end of the each spring is fixed on a rigid support and the other end is free as shown fig. The distance CD between the free ends of a spring is 60cm.

## Springs in Series - University of Iceland

Two springs in series Consider two massless springs connected in series. Spring 1 has a spring constant , and spring 2 has a spring constant . A constant force of magnitude is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a

## 2 springs,with force constants k1=125N/m and k2=275N/m are

Mar 29, 2017  2 springs,with force constants k1=125N/m and k2=275N/m are connected.Mass=0.95kg is attached to the springs,what is the amount of stretch,x?

## Period of mass with two springs - HELP please? - Yahoo Answers

Jul 23, 2011  A mass m slides on a frictionless horizontal surface, connected to two springs. If the springs have force constants k1 and k2. Show that the simple harmonic motion has period: T = 2pi * sqrt {m [(k1+k2)/(k1*k2)]} I can't seem to figure it out, please help!

## Hooke's Law, Examples with solutions - Physics

Springs in Parallel Two springs are said to be in parallel when used as in the figure below. The two springs behave like on spring whose constant k is given by k = k1 + k2 Problem 3 What is the magnitude of the force required to stretch two springs of constants k1 = 100 N / m and k2 = 200 N / m by 6 cm if they are in parallel? Solution

## Work power-energy - SlideShare

Aug 22, 2014  Two springs have their force constant as K1 and K2 (K1 > K2). When they are stretched by the same force : A) no work is done in case of both the springs B) equal work is done in case of both the springs C) more work is done in case of second spring D) more work is done in case of first spring 18.

## 3 Springs and 2 Masses - Physics Forums

Dec 02, 2018  Two masses (M1 and M2) are connected together by 3 springs. The spring constants are k1, k2, k3. Block 1 is at equilibrium at x=0. Block 2 is at equilibrium at x=1. Determine a function for the force on the blocks. 2. Homework Equations 3. The Attempt at a Solution Without using the initial values, I was able to get the following equations

## Two blocks of mass 'm' are connected to a spring of spring

Oct 11, 2019  On release of masses, the half-springs experience restoring force and the masses move. With the result the mid-point of spring remains stationary and both half-spring-mass systems execute SHM, in opposite directions. Each sub-system, has initial PE =(1/4)kx^2. Spring constant of each half-spring = 2k. Initial compression is also shared equally = x/2.

## Two springs are in a series combination and are attached

Jun 26, 2018  Two springs are in a series combination and are attached to a block of mass 'm' which is in equilibrium. Spring constants:k and k’ Extensions:x and x’ respectively. Find the force exerted by spring

## What is the spring constant in parallel connection and

Dec 15, 2016  Parallel. When two massless springs following Hooke's Law, are connected via a thin, vertical rod as shown in the figure below, these are said to be connected in parallel. Spring 1 and 2 have spring constants k_1 and k_2 respectively. A constant force vecF is exerted on the rod so that remains perpendicular to the direction of the force. So that the springs are extended by the same amount.

## If 2 springs of spring constants k1 k2 are connected in

If 2 springs of spring constants k1, k2 are connected in series, the equivalent spring constant would be k, defined by 1 / k = = 1 / k1 + 1 / k2 Thus 1 / k > 1 / k1 and 1 / k > 1 / k2; hence k < k1 and k < k2 Experiment 4(Parallel) Trial Weight (kg) Force(N) Spring Constant Displacement(m) 1.1 0.980 980N/m.001 2.15 1.47 294N/m.005 3.25 2.45 122.5N/m.020 4.35 3.43 98N/m.035 5.55 5.39 85.6N/m.063

## 2 springs,with force constants k1=125N/m and k2=275N/m are

Mar 29, 2017  2 springs,with force constants k1=125N/m and k2=275N/m are connected.Mass=0.95kg is attached to the springs,what is the amount of stretch,x?

## A block of mass m is connected to two springs of force

Answer to: A block of mass m is connected to two springs of force constants k1 and k2. The block moves on a frictionless table after it is

## What is the equivalent spring constant when springs are

We get F = x*(k1 + k2) = keq*x. The equivalent spring constant keq = k1 + k2. If two springs are connected in series, and a force F is applied, the force acts on both the springs changing the length of each differently. Let the spring constants of the springs be k1 and k2 and the change in length be x1 and x2.

## Two springs of constants k1 and k2 are connected in series

Dec 19, 2011  Two springs of constants k1 and k2 are connected in series to a block a that vibrates in simple harmonic motion with a period of 5 s. When the same

## Questions : Work by variable force - Unacademy

Two springs have their force constant as k and k2(ki >k2). When they are stretched by the same force (a) No work is done in case of both the springs (b) Equal work is done in case of both the springs (c) More work is done in case of second spring (d) More work is done in case of first spring 4.A spring of force constant 10 Nm has an initial stretch 0.20 m.

## What's the equation that relates? - Yahoo Answers

Apr 01, 2007  Best Answer: The answer depends on whether the springs are connected in series or in parallel: (1) in series: (1/k total) = (1/k1) + (1/k2) (2) in parallel: k total = k1 + k2. Source(s):

## Worked example 11.3: Block and two springs

Given that the force constants of the two springs are and , find the period of oscillation of the system. Answer: Let and represent the extensions of the first and second springs, respectively. Let and be the magnitudes of the forces exerted by the first and second springs, respectively.

## What is parallel spring system and series spring system

Here, both the springs have a different constant or stiffness say k1 and k2. Thus, when one moves the mass by a displacement of x, one spring is pushed which pushes the other spring as well. Their combined displacement is x but we are yet to find their individual displacements.

## Springs--Two Springs and a Mass -- from Eric Weisstein's

Springs--Two Springs and a Mass. Consider a mass m with a spring on either end, each attached to a wall. Let and be the spring constants of the springs. A displacement of the mass by a distance x results in the first spring lengthening by a distance x (and pulling in the direction), while the second spring is compressed by a

## PC Advanced Physics Final Flashcards - Quizlet

Two springs are identical except for their force constants, k2>k1. If the same force is used to stretch the springs, compare the distance each spring will be stretched. When stretching, k1 will stretch a greater distance b/c it is easier to pull than k2.

## LABORATORY VIII MECHANICAL OSCILLATIONS

You have two different springs that have the same unstretched length, but different spring constants k1 and k2. These springs can be hung vertically side-by-side (Figure A) or end-to-end (Figure B).

## How to solve this type of physics problems - Quora

When two springs with spring constant or stiffness k1 & k2 are attached in series then the resultant spring constant K is equal to the sum of the reciprocal of the individual spring constant i.e 1/K =

## (a) Two springs with spring constants k1 and k2 are

Jul 25, 2018  (a) Two springs with spring constants k 1 and k 2 are connected in parallel, as shown in Fig. 4.18.What is the effective spring constant, k eff? In other words, if the mass is displaced by x, find the k eff for which the force equals F = -k eff x.


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Aerospace Wave Propagation solution for a variable area 1D duct

1. Sep 26, 2011

domainxh

In the problem I am suppose to use the wave equation to solve it.

I assume 1D plane wave duct,

u(x,t) = 1/(rho*C)*real((Aexp(ikx)-Bexp(-ikx))exp(iwt))
where C is the speed of sound, u is the velocity, p is the pressure, w is the angular frequency, t is time, rho is the density, and both A and B are unknown constants which I have to find.

The equation for Pressure is similar except without density and speed of sound term,
p(x,t) = real((Aexp(ikx)-Bexp(-ikx))exp(iwt))

At the area changing point, I am suppose to assume the same pressure and Aa*U1 = Ab*U2 (essentially matching the conditions)

After applying the left end B.C, I get A = B
But I can't move any further.... Any help would be greatly appreciated.

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61.4 KB
Views:
70

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text:
Volume 385 - 10th International Workshop on Charm Physics (CHARM2020) - Parallel programme
Charmonium in nuclear matter and nuclei
J. Cobos Martinez*, K. Tsushima, G. Krein and A.W. Thomas
Full text: pdf
Pre-published on: September 22, 2021
Published on: November 10, 2021
Abstract
We present results for the $\eta_c$-nucleus bound state energies for various nuclei using
an effective Lagrangians approach. The attractive potentials for the $\eta_c$ in the nuclear medium
originate from the medium-modified intermediate $D D^{*}$ state in the $\eta_c$ self energy,
using the local density approximation. Our results suggest that the $\eta_c$ should form bound states with all the nuclei considered.
DOI: https://doi.org/10.22323/1.385.0041
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
## Pacific Journal of Mathematics

### A remark on functions of bounded mean oscillation and bounded harmonic functions. Addendum to: BMO functions and the $\overline\partial$-equation''.

N. Th. Varopoulos

#### Article information

Source
Pacific J. Math., Volume 74, Number 1 (1978), 257-259.

Dates
First available in Project Euclid: 8 December 2004

https://projecteuclid.org/euclid.pjm/1102810455

Mathematical Reviews number (MathSciNet)
MR0508036

Zentralblatt MATH identifier
0382.31004

Subjects
Primary: 32A99: None of the above, but in this section
Secondary: 46E99: None of the above, but in this section

#### Citation

Varopoulos, N. Th. A remark on functions of bounded mean oscillation and bounded harmonic functions. Addendum to: BMO functions and the $\overline\partial$-equation''. Pacific J. Math. 74 (1978), no. 1, 257--259. https://projecteuclid.org/euclid.pjm/1102810455

#### References

• [1] L. Carleson, Interpolation by bounded analytic functionsand the Corona problem, Ann. of Math., 76 (1962), 547-559.
• [2] C. FefFerman and E. M. Stein, Hp spaces of several variables, Acta Math., 129 (1972), 137-193.
• [3] E. Galiardo, Caratterizzazioni delta trace sulla frontier a relative ad alcune classi di funzioniin n variabili, Rend. Sem. Mat. Univ. Padova, 27 (1957), 284-305.
• [4] D. Marshall, Subalgebras of Z/ containing H, Acta Math., 137, 91-98.
• [5] W. Rudin, The radial variation of analytic functions,Duke Math. J., 22 (1955), 235-242.
• [6] E. M. Stein, SingularIntegralsand DifferentiabilityProperties ofFunctions, Princeton Univ. Press (1970).
• [7] N. Th. Varopoulos, BMO functionsand the d equation,Pacific J. Math., 71 (1) (1977), 221-273.

• Orig : N. Th. Varopoulos. BMO functions and the $\overline\partial$-equation. Pacific Journal of Mathematics volume 71, issue 1, (1977), pp. 221-273.


text:
# Properties

 Label 161700di Number of curves $2$ Conductor $161700$ CM no Rank $0$ Graph

# Related objects

Show commands for: SageMath
sage: E = EllipticCurve("di1")

sage: E.isogeny_class()

## Elliptic curves in class 161700di

sage: E.isogeny_class().curves

LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
161700.bo2 161700di1 $$[0, -1, 0, -26133, -5090238]$$ $$-67108864/343035$$ $$-10089431178750000$$ $$[2]$$ $$1105920$$ $$1.7552$$ $$\Gamma_0(N)$$-optimal
161700.bo1 161700di2 $$[0, -1, 0, -632508, -193066488]$$ $$59466754384/121275$$ $$57071529900000000$$ $$[2]$$ $$2211840$$ $$2.1017$$

## Rank

sage: E.rank()

The elliptic curves in class 161700di have rank $$0$$.

## Complex multiplication

The elliptic curves in class 161700di do not have complex multiplication.

## Modular form 161700.2.a.di

sage: E.q_eigenform(10)

$$q - q^{3} + q^{9} + q^{11} - 6q^{13} + 2q^{17} + O(q^{20})$$

## Isogeny matrix

sage: E.isogeny_class().matrix()

The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the Cremona numbering.

$$\left(\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right)$$

## Isogeny graph

sage: E.isogeny_graph().plot(edge_labels=True)

The vertices are labelled with Cremona labels.


text:
An animation demonstrating the relationship between volume and temperature.

Charles's law (also known as the law of volumes) is an experimental gas law which describes how gases tend to expand when heated. A modern statement of Charles's law is:

The volume of a given mass of an ideal gas is directly proportional to its temperature on the absolute temperature scale (in Kelvins) if pressure and the amount of gas remain constant; that is, the volume of the gas increases or decreases by the same factor as its temperature.[1]

which can be written as:

 $V \propto T\,$

or

 $\frac{V}{T}=k$

where:

V is the volume of the gas
T is the temperature of the gas (measured in Kelvin).
k is a constant.

This law explains how a gas expands as the temperature increases; conversely, a decrease in temperature will lead to a decrease in volume. For comparing the same substance under two different sets of conditions, the law can be written as:

 $\frac{V_1}{T_1} = \frac{V_2}{T_2} \qquad \mathrm{or} \qquad \frac {V_2}{V_1} = \frac{T_2}{T_1} \qquad \mathrm{or} \qquad V_1 T_2 = V_2 T_1.$

The equation shows that, as absolute temperature increases, the volume of the gas also increases in proportion. The law was named after scientist Jacques Charles, who formulated the original law in his unpublished work from the 1780s.

## History

It was first published by French natural philosopher Joseph Louis Gay-Lussac in 1802,[2] although he credited the discovery to unpublished work from the 1780s by Jacques Charles. The law was independently discovered by British natural philosopher John Dalton by 1801, although Dalton's description was less thorough than Gay-Lussac's.[3][4] The basic principles had already been described a century earlier by Guillaume Amontons.[5]

Gay-Lussac was the first to demonstrate that the law applied generally to all gases, and to the vapours of volatile liquids if the temperature was more than a few degrees above the boiling point.[6][citation needed] His statement of the law can be expressed mathematically as:

 $V_{100} - V_0 = kV_0\,$

where V100 is the volume occupied by a given sample of gas at 100 °C; V0 is the volume occupied by the same sample of gas at 0 °C; and k is a constant which is the same for all gases at constant pressure. Gay-Lussac's value for k was 12.6666, remarkably close to the present-day value of 12.7315. This law was first given by J. Charles in 1787.

## Relation to the ideal gas law

French physicist Émile Clapeyron combined Charles's law with Boyle's law in 1834 to produce a single statement which would become known as the ideal gas law.[7] Claypeyron's original statement was:

 $pV = \frac{p_0 V_0}{267 + t_0} (267 + t)$

where t is the Celsius temperature; and p0, V0 and t0 are the pressure, volume and temperature of a sample of gas under some standard state. The figure of 267 came directly from Gay-Lussac's work: the modern figure would be 273.15. For any given sample of gas, p0V0267+t0 is a constant (Clapeyron denoted this constant R, and it is closely related to the modern gas constant); if the pressure is also constant, the equation simplifies to

 $V \propto (267 + t)\,$

as required.

The modern statement of the ideal gas law is:

 $pV = nRT\,$

where n is the amount of substance of the gas sample; and R is the gas constant. The amount of substance is constant for any given gas sample so, at constant pressure, the equation rearranges to:

 $V = \frac{nR}{p} T$

where nRp is the constant of proportionality.

An ideal gas is defined as a gas which obeys the ideal gas law, so Charles's law is only expected to be followed exactly by ideal gases. Nevertheless, it is a good approximation to the behaviour of real gases at relatively high temperatures and relatively low pressures.

## Relation to absolute zero

Charles's law appears to imply that the volume of a gas will descend to zero at a certain temperature (−266.66 °C according to Gay-Lussac's figures) or -273°C. Gay-Lussac was clear in his description that the law was not applicable at low temperatures:

but I may mention that this last conclusion cannot be true except so long as the compressed vapors remain entirely in the elastic state; and this requires that their temperature shall be sufficiently elevated to enable them to resist the pressure which tends to make them assume the liquid state.[2]

Gay-Lussac had no experience of liquid air (first prepared in 1877), although he appears to believe (as did Dalton) that the "permanent gases" such as air and hydrogen could be liquified. Gay-Lussac had also worked with the vapours of volatile liquids in demonstrating Charles's law, and was aware that the law does not apply just above the boiling point of the liquid:

I may however remark that when the temperature of the ether is only a little above its boiling point, its condensation is a little more rapid than that of atmospheric air. This fact is related to a phenomenon which is exhibited by a great many bodies when passing from the liquid to the solid state, but which is no longer sensible at temperatures a few degrees above that at which the transition occurs.[2]

The first mention of a temperature at which the volume of a gas might descend to zero was by William Thomson (later known as Lord Kelvin) in 1848:[8]

This is what we might anticipate, when we reflect that infinite cold must correspond to a finite number of degrees of the air-thermometer below zero; since if we push the strict principle of graduation, stated above, sufficiently far, we should arrive at a point corresponding to the volume of air being reduced to nothing, which would be marked as −273° of the scale (−100/.366, if .366 be the coefficient of expansion); and therefore −273° of the air-thermometer is a point which cannot be reached at any finite temperature, however low.

However, the "absolute zero" on the Kelvin temperature scale was originally defined in terms of the second law of thermodynamics, which Thomson himself described in 1852.[9] Thomson did not assume that this was equal to the "zero-volume point" of Charles's law, merely that Charles's law provided the minimum temperature which could be attained. The two can be shown to be equivalent by Ludwig Boltzmann's statistical view of entropy (1870).

## Relation to kinetic theory

The kinetic theory of gases relates the macroscopic properties of gases, such as pressure and volume, to the microscopic properties of the molecules which make up the gas, particularly the mass and speed of the molecules. In order to derive Charles's law from kinetic theory, it is necessary to have a microscopic definition of temperature: this can be conveniently taken as the temperature being proportional to the average kinetic energy of the gas molecules, Ek:

 $T \propto \bar{E_{\rm k}}.\,$

Under this definition, the demonstration of Charles's law is almost trivial. The kinetic theory equivalent of the ideal gas law relates pV to the average kinetic energy:

 $pV = \frac{2}{3} N \bar{E_{\rm k}}\,$

where N is the number of molecules in the gas sample. If the pressure is constant, the volume is directly proportional to the average kinetic energy (and hence to the temperature) for any given gas sample.

## References

1. ^ Fullick, P. (1994), Physics, Heinemann, pp. 141–42, ISBN 0-435-57078-1.
2. ^ a b c Gay-Lussac, J. L. (L'An X – 1802), "Recherches sur la dilatation des gaz et des vapeurs" [Researches on the expansion of gases and vapors], Annales de chimie 43: 137–175. English translation (extract).
On page 157, Gay-Lussac mentions the unpublished findings of Charles: "Avant d'aller plus loin, je dois prévenir que quoique j'eusse reconnu un grand nombre de fois que les gaz oxigène, azote, hydrogène et acide carbonique, et l'air atmosphérique se dilatent également depuis 0° jusqu'a 80°, le cit. Charles avait remarqué depuis 15 ans la même propriété dans ces gaz ; mais n'avant jamais publié ses résultats, c'est par le plus grand hasard que je les ai connus." (Before going further, I should inform [you] that although I had recognized many times that the gases oxygen, nitrogen, hydrogen, and carbonic acid [i.e., carbon dioxide], and atmospheric air also expand from 0° to 80°, citizen Charles had noticed 15 years ago the same property in these gases; but having never published his results, it is by the merest chance that I knew of them.)
3. ^ J. Dalton (1802) "Essay IV. On the expansion of elastic fluids by heat," Memoirs of the Literary and Philosophical Society of Manchester, vol. 5, pt. 2, pages 595-602.
4. ^ http://www.chemistryexplained.com/Fe-Ge/Gay-Lussac-Joseph-Louis.html
5. ^ See:
6. ^ Gay-Lussac (1802), from page 166:
"Si l'on divise l'augmentation totale de volume par le nombre de degrés qui l'ont produite ou par 80, on trouvera, en faisant le volume à la température 0 égal à l'unité, que l'augmentation de volume pour chaque degré est de 1 / 223.33 ou bien de 1 / 266.66 pour chaque degré du thermomètre centrigrade."
If one divides the total increase in volume by the number of degrees that produce it or by 80, one will find, by making the volume at the temperature 0 equal to unity (1), that the increase in volume for each degree is 1 / 223.33 or 1 / 266.66 for each degree of the centigrade thermometer.
From page 174:
" … elle nous porte, par conséquent, à conclure que tous les gaz et toutes les vapeurs se dilatent également par les mêmes degrés de chaleur."
… it leads us, consequently, to conclude that all gases and all vapors expand equally [when subjected to] the same degrees of heat.
7. ^ Clapeyron, E. (1834), "Mémoire sur la puissance motrice de la chaleur", Journal de l'École Polytechnique XIV: 153–90. Facsimile at the Bibliothèque nationale de France (pp. 153–90).
8. ^ .
9. ^ Thomson, William (1852), "On the Dynamical Theory of Heat, with numerical results deduced from Mr Joule's equivalent of a Thermal Unit, and M. Regnault's Observations on Steam", Philosophical Magazine 4. Extract.


text:
# Sandblasting cabin for the railway sector

We have carried out a large project for a multinational leader in the railway field: a manual shot blasting cabin 31 meters long, 6 meters high and 6 meters wide for the sandblasting of railway carriages; equipped with pit for underbody sandblasting, total recovery system of the grit through scrapers, auger, bucket elevator, separator, silos and two dust collecting filters, also equipped with two platforms for lifting the operators and a compressor that simultaneously powers four sandblasters.

Tags:


text:
To install click the Add extension button. That's it.

The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.

4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
.
Leo
Newton
Brights
Milds

# Graveyard orbit

Orbit size comparison of GPS, GLONASS, Galileo, BeiDou-2, and Iridium constellations, the International Space Station, the Hubble Space Telescope, and geostationary orbit (and its graveyard orbit), with the Van Allen radiation belts and the Earth to scale.[a]
The Moon's orbit is around 9 times as large as geostationary orbit.[b] (In the SVG file, hover over an orbit or its label to highlight it; click to load its article.)

A graveyard orbit, also called a junk orbit or disposal orbit, is an orbit that lies away from common operational orbits. One significant graveyard orbit is a supersynchronous orbit well above geosynchronous orbit. Some satellites are moved into such orbits at the end of their operational life to reduce the probability of colliding with operational spacecraft and generating space debris.

## Overview

A graveyard orbit is used when the change in velocity required to perform a de-orbit maneuver is too large. De-orbiting a geostationary satellite requires a delta-v of about 1,500 metres per second (4,900 ft/s), whereas re-orbiting it to a graveyard orbit only requires about 11 metres per second (36 ft/s).[1]

For satellites in geostationary orbit and geosynchronous orbits, the graveyard orbit is a few hundred kilometers above the operational orbit. The transfer to a graveyard orbit above geostationary orbit requires the same amount of fuel as a satellite needs for about three months of stationkeeping. It also requires a reliable attitude control during the transfer maneuver. While most satellite operators plan to perform such a maneuver at the end of their satellites' operational lives, through 2005 only about one-third succeeded.[2] However, as of 2011, most[clarification needed] recently decommissioned geosynchronous spacecraft were said to have been moved to a graveyard orbit.[3]

According to the Inter-Agency Space Debris Coordination Committee (IADC)[4] the minimum perigee altitude ${\displaystyle \Delta {H}\,}$ above the geostationary orbit is:

${\displaystyle \Delta {H}=235{\mbox{ km}}+\left(1000C_{R}{\frac {A}{m}}\right){\mbox{ km}}}$

where ${\displaystyle C_{R}\,}$ is the solar radiation pressure coefficient and ${\displaystyle {\frac {A}{m}}\,}$ is the aspect area [m2] to mass [kg] ratio of the satellite. This formula includes about 200 km for the GEO-protected zone to also permit orbit maneuvers in GEO without interference with the graveyard orbit. Another 35 kilometres (22 mi) of tolerance must be allowed for the effects of gravitational perturbations (primarily solar and lunar). The remaining part of the equation considers the effects of the solar radiation pressure, which depends on the physical parameters of the satellite.

In order to obtain a license to provide telecommunications services in the United States, the Federal Communications Commission (FCC) requires all geostationary satellites launched after March 18, 2002, to commit to moving to a graveyard orbit at the end of their operational life.[5] U.S. government regulations require a boost, ${\displaystyle \Delta {H}}$, of about 300 km.[6]

A spacecraft moved to a graveyard orbit will typically be passivated.

Uncontrolled objects in a near geostationary [Earth] orbit (GEO) exhibit a 53-year cycle of orbital inclination[7] due to the interaction of the Earth's tilt with the lunar orbit. The orbital inclination varies ± 7.4°, at up to 0.8°pa.[7]:3

## Disposal orbit

While the standard geosynchronous satellite graveyard orbit results in an expected orbital lifetime of millions of years, the increasing number of satellites, the launch of microsatellites, and the FCC approval of large megaconstellations of thousands of satellites for launch by 2022 necessitates new approaches for deorbiting to assure earlier removal of the objects once they have reached end-of-life. Contrary to GEO graveyard orbits requiring three months' worth of fuel, large satellite networks require orbits that passively decay into the Earth's atmosphere. For example, both OneWeb and SpaceX have committed to the FCC regulatory authorities that decommissioned satellites will decay to a lower orbit — a disposal orbit—where the satellite orbital altitude would decay due to atmospheric drag and then naturally reenter the atmosphere and burn up within one year of end-of-life.[8]

## Notes

1. ^ Orbital periods and speeds are calculated using the relations 4π2R3 = T2GM and V2R = GM, where R, radius of orbit in metres; T, orbital period in seconds; V, orbital speed in m/s; G, gravitational constant, approximately 6.673×10−11 Nm2/kg2; M, mass of Earth, approximately 5.98×1024 kg.
2. ^ Approximately 8.6 times (in radius and length) when the moon is nearest (363104 km ÷ 42164 km) to 9.6 times when the moon is farthest (405696 km ÷ 42164 km).

## References

1. ^ "Method for re-orbiting a dual-mode propulsion geostationary spacecraft – Patent # 5651515 – PatentGenius". Archived from the original on 2013-11-10. Retrieved 2012-10-28.
2. ^
3. ^ Johnson, Nicholas (2011-12-05). Livingston, David (ed.). "Broadcast 1666 (Special Edition) – Topic: Space debris issues" (podcast). The Space Show. 1:03:05-1:06:20. Retrieved 2015-01-05.
4. ^ "Archived copy" (PDF). Archived from the original (PDF) on 2015-04-02. Retrieved 2015-03-07.CS1 maint: archived copy as title (link)
5. ^ "FCC Enters Orbital Debris Debate". Archived from the original on March 8, 2005.
6. ^
7. ^ a b Anderson, Paul; et al. (2015). Operational Considerations of GEO Debris Synchronization Dynamics (PDF). 66th International Astronautical Congress. Jerusalem, Israel. IAC-15,A6,7,3,x27478.
8. ^ Brodkin, Jon (4 October 2017). "SpaceX and OneWeb broadband satellites raise fears about space debris". Ars Technica. Retrieved 28 April 2019.


text:
New Titles  |  FAQ  |  Keep Informed  |  Review Cart  |  Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education

Thirteen Papers Translated from the Russian
 SEARCH THIS BOOK:
American Mathematical Society Translations--Series 2
1989; 140 pp; hardcover
Volume: 144
ISBN-10: 0-8218-3125-9
ISBN-13: 978-0-8218-3125-0
List Price: US$86 Member Price: US$68.80
Order Code: TRANS2/144

Topics include group representations, $$p$$-classes, approximation of functions, entire functions, the Laplace operator, and von Neumann algebras

• S. D. Berman and K. Buzasi -- On modules over group algebras of groups containing an infinite cyclic subgroup of finite index
• S. D. Berman and K. Buzasi -- Description of all finite-dimensional real representations of groups containing an infinite cyclic subgroup of finite index
• I. S. Kats -- Connection between integral characteristics of growth of entire functions and distributions of their zeros
• Yu. I. Lyubarskiĭ -- Representation of functions from $$H^p$$ in a half-plane, and some of its applications
• V. K. Dubovoĭ -- Indefinite metric in Schur's interpolation problem for analytic functions. I
• V. K. Dubovoĭ -- Indefinite metric in Schur's interpolation problem for analytic functions. II
• A. A. Nersesyan -- On uniform and tangential approximation by meromorphic functions
• S. Yu. Favorov -- On entire functions of completely regular growth of several variables
• V. E. Lyantse and Kh. B. Maiorga -- On the theory of the one-point boundary value problem for the Laplace operator. I
• V. E. Lyantse and Kh. B. Maiorga -- On the theory of the one-point boundary value problem for the Laplace operator. II
• G. N. Zholtkevich -- On the cohomology of von Neumann algebras
• N. A. Pustovoĭtov -- On approximate algebraic stability criteria, numerical-analytical methods for investigating the dynamics and stability of complex systems
• I. V. Lobarev -- Eigenvalues and eigenfunctions of the Peierls equation


text:
emis estimates vehicular emissions as the product of the vehicles on a road, length of the road, emission factor avaliated at the respective speed. $$E = VEH*LENGTH*EF(speed)$$

emis(veh, lkm, ef, speed, agemax = ifelse(is.data.frame(veh), ncol(veh),
ncol(veh[[1]])), profile, simplify = FALSE, hour = nrow(profile),
day = ncol(profile), array = TRUE, verbose = FALSE)

## Arguments

veh "Vehicles" data-frame or list of "Vehicles" data-frame. Each data-frame as number of columns matching the age distribution of that ype of vehicle. The number of rows is equal to the number of streets link. If this is a list, the length of the list is the vehicles for each hour. Length of each link in km List of functions of emission factors Speed data-frame with number of columns as hours. The default value is 34km/h Age of oldest vehicles for that category Dataframe or Matrix with nrows equal to 24 and ncol 7 day of the week Logical; to determine if EmissionsArray should les dimensions, being streets, vehicle categories and hours or default (streets, vehicle categories, hours and days). Default is FALSE to avoid break old code, but the recommendation is that new estimations use this parameter as TRUE Number of considered hours in estimation. Default value is number of rows of argument profile Number of considered days in estimation Deprecated! emis_cold returns only arrays. When TRUE and veh is not a list, expects a profile as a dataframe producing an array with dimensions (streets x columns x hours x days) Logical; To show more information

## Value

If the user applies a top-down approach, the resulting units will be according its own data. For instance, if the vehicles are veh/day, the units of the emissions implicitly will be g/day.

## Note

Hour and day will be deprecated because they can be infered from the profile matrix.

## Examples

if (FALSE) {
# Do not run
data(net)
data(pc_profile)
data(profiles)
data(fe2015)
data(fkm)
PC_G <- c(33491,22340,24818,31808,46458,28574,24856,28972,37818,49050,87923,
133833,138441,142682,171029,151048,115228,98664,126444,101027,
84771,55864,36306,21079,20138,17439, 7854,2215,656,1262,476,512,
1181, 4991, 3711, 5653, 7039, 5839, 4257,3824, 3068)
veh <- data.frame(PC_G = PC_G)
pc1 <- my_age(x = net$ldv, y = PC_G, name = "PC") # Estimation for morning rush hour and local emission factors speed <- data.frame(S8 = net$ps)
lef <- EmissionFactorsList(fe2015[fe2015$Pollutant=="CO", "PC_G"]) E_CO <- emis(veh = pc1,lkm = net$lkm, ef = lef, speed = speed,
profile = 1)

# Estimation for 168 hour and local factors
pcw <- temp_fact(net$ldv+net$hdv, pc_profile)
speed <- netspeed(pcw, net$ps, net$ffs, net$capacity, net$lkm, alpha = 1)
lef <- EmissionFactorsList(fe2015[fe2015$Pollutant=="CO", "PC_G"]) E_CO <- emis(veh = pc1, lkm = net$lkm,
ef = lef,
speed = speed,
profile = profiles$PC_JUNE_2014) summary(E_CO) lpc <- list(pc1, pc1) E_COv2 <- emis(veh = lpc,lkm = net$lkm, ef = lef, speed = speed)
# top down
veh <- age_ldv(x = net$ldv[1:4], name = "PC_E25_1400", agemax = 4) mil <- fkm$KM_PC_E25(1:4)
ef <- ef_cetesb("COd", "PC_G")[1:4]
emis(veh, units::set_units(mil, "km"), ef)
# group online
bus1 <- age_hdv(30, agemax = 4)
veh = bus1
lkm = units::set_units(400, "km")
speed = 40
efco <- ef_cetesb("COd", "UB", agemax = 4)
lef <- ef_hdv_scaled(dfcol = as.numeric(efco),
v = "Ubus",
t = "Std",
g = ">15 & <=18",
eu = rep("IV", 4),
gr = 0,
l = 0.5,
p = "CO")
for(i in 1:length(lef)) print(lef[[i]](10))
emis(veh = bus1, lkm = lkm, speed = 40, ef = lef, verbose = T)
emis(veh = bus1, lkm = lkm, ef = efco, verbose = T)
}


text:
How to swap two files in Linux command line?

As system administrators or DevOps functions, many times we come with the requirement to exchange files content, for example, suppose you have a backup file of /etc/passwd by name /etc/password.backup and you want to restore the same in /etc/passwd but you also want current content of /etc/password should be copied to /etc/passwd.backup. In other words, exchange content of /etc/passwd with /etc/passwd.backup and /etc/passwd.backup with /etc/passwd.

The Linux operating system's useful tools and commands enable you to achieve/accomplish a wide range of file manipulation goals. For one reason or another, you may need to swap two files in the Linux file system.

When we talk about swapping two files on a Linux operating system, we don't mean swapping/exchanging the location of these two files, but rather their actual content. In this article, we will understand how to swap two files in Linux command line.

Consider the existence of the following files in a Linux operating system environment to better understand the goal of this tutorial.

Create two files f1.txt and f2.txt.

File f1.txt has the following contents

Hi Welcome in Linux

File f2.txt has the following contents

Hi Welocome in Linux shell scripting
[sachin@lmdesk swapping]$vim f1.txt [sachin@lmdesk swapping]$ vim f2.txt
[sachin@lmdesk swapping]$cat f1.txt Hi Welcome in Linux! [sachin@lmdesk swapping]$ cat f2.txt
Hi Welcome in Linux shell scripting

If we successfully swap these two files

File f1.txt will read − Hi Welocome in Linux shell scripting

File f2.txt will read − Hi Welcome in Linux

Swapping contents of two files by using mv command

The mv command can be used to rename files and can also be used to move them from one location to another. Here the purpose is to swap two files in the command line, so we using mv command for renaming purposes, not for moving files to other locations.

Exchanging values between variables is a common requirement in programming and scripting jobs. When we want to exchange the values of two variables, in most programming languages, the best approach is to use a temp variable. Examine the following demonstration

temp = x
x = y
y = temp

Lets implement this algorithm with mv command, mv command will be modified like this

mv f1.txt test.txt
mv f2.txt f1.txt
mv test.txt f2.txt

The output of these command will swap two files.

Swapping contents of two files by using shell script

if (( $# == 2)) ; then TMPFILE=$(mktemp $(dirname "$1")/function.txt)
mv "$1"$TMPFILE && mv "$2" "$1" && mv $TMPFILE "$2"
else
echo "Error: Two valid file paths required"
return 1
fi

Let us understand above shell script. \$# is the number of arguments to shell script, so first line of script with if statement validating if the argument to shell script is 2 then and only then exchange contents. TMPFILE is a variable holding path of file which hold content temporarily in the process of content exchange, mv command we discussed earlier and here we using same for renaming purpose. If two files are not passed to this script, then error will get generated

We can simply pass the two files to swap as arguments. The function will handle everything else. Bash's auto-completion will come in handy here.

Conclusion

In this article, we learned how to swap the contents of two files using three mv commands. We've also written a shell script to make it easier to carry out this operation. You can either use the command execution approach or can use shell script approach.


text:
# Defining a function as a truncated taylors series

I am trying to numerically find the error of a truncated series. To do so I want to first define a function that is the truncated sum, I cannot seem to do that. All I am tried is not working.

``f[x_]:=Normal[Series[ArcTan[x], {x, 0, 100}]] ``

Does not work as f[1] evaluates $$x$$ inside $$\arctan$$. I also tried

``Normal[Series[ArcTan[x], {x, 0, 100}]] f[x_]:=% ``

But that did not work. When I copy and paste the result of

``Normal[Series[ArcTan[x], {x, 0, 100}]] ``

It works fine. What am I doing wrong?


text:
# strange unicode behavior in listings [duplicate]

I have the minimal document here:

\documentclass{article}
\usepackage{unicode-math}
\usepackage{listings}

\setmainfont{Stix}
\setmathfont{Latin Modern Math}
\setmonofont{Fira Mono}

\begin{document}
this is λ lambda
$$\int_0^\infty λ\lambda$$
\begin{lstlisting}
# some random pseudocode using unicode
def function(λ):
αδ = βθ

def function( λ):
pass
\end{lstlisting}
\end{document}

producing with lualatex:

Look at how the unicode characters are rendered in the listing, but not in the right place.

I'm afraid you need to add manually all the Unicode characters you need.

\documentclass{article}
\usepackage{unicode-math}
\usepackage{listings}

\setmainfont{STIX Two Text}
\setmathfont{STIX Two Math}
\setmonofont{Fira Mono}

\makeatletter % see https://tex.stackexchange.com/a/320345
\lst@InputCatcodes
\def\lst@DefEC{%
\lst@CCECUse \lst@ProcessLetter
^^80^^81^^82^^83^^84^^85^^86^^87^^88^^89^^8a^^8b^^8c^^8d^^8e^^8f%
^^90^^91^^92^^93^^94^^95^^96^^97^^98^^99^^9a^^9b^^9c^^9d^^9e^^9f%
^^b0^^b1^^b2^^b3^^b4^^b5^^b6^^b7^^b8^^b9^^ba^^bb^^bc^^bd^^be^^bf%
^^c0^^c1^^c2^^c3^^c4^^c5^^c6^^c7^^c8^^c9^^ca^^cb^^cc^^cd^^ce^^cf%
^^d0^^d1^^d2^^d3^^d4^^d5^^d6^^d7^^d8^^d9^^da^^db^^dc^^dd^^de^^df%
^^e0^^e1^^e2^^e3^^e4^^e5^^e6^^e7^^e8^^e9^^ea^^eb^^ec^^ed^^ee^^ef%
^^f0^^f1^^f2^^f3^^f4^^f5^^f6^^f7^^f8^^f9^^fa^^fb^^fc^^fd^^fe^^ff%
^^^^20ac^^^^0153^^^^0152%
^^^^03b1% alpha
^^^^03b2% beta
^^^^03b4% delta
^^^^03bb% lambda
^^^^03b8% theta
^^00}
\lst@RestoreCatcodes
\makeatother

\lstset{basicstyle=\ttfamily}

\begin{document}
this is λ lambda
$\int_0^\infty λ\lambda$
\begin{lstlisting}
# some random pseudocode using unicode
def function(λ):
αδ = βθ

def function(λ):
pass
\end{lstlisting}
\end{document}

As an alternative you can use the minted package (compile with --shell-escape). Note that I changed the fonts because I don't have STIX or Fira Mono.

\documentclass{article}
\usepackage{unicode-math}
\usepackage{minted}

\setmainfont{CMU Serif}
\setsansfont{CMU Sans Serif}
\setmonofont{CMU Typewriter Text}
\setmathfont{Latin Modern Math}

\begin{document}
this is λ lambda
$$\int_0^\infty λ\lambda$$
\begin{minted}{python}
# some random pseudocode using unicode
def function(λ):
αδ = βθ

def function( λ):
pass
\end{minted}
\end{document}

• It does not answer the question which was about listings. It works well with minted, even the unicode characters appear with a red rectangle around. – Hugo Trentesaux Dec 2 '20 at 10:47
• @HugoTrentesaux I agree that it does not answer the question as asked, but it may be a useful alternative (for you and/or for future visitors of the site). As you can see I don't have the red rectangles in my output (and indeed this is valid Python code). – Marijn Dec 2 '20 at 10:54
• Maybe tex.stackexchange.com/questions/343494/… could be useful to address the red rectangles, see also tex.stackexchange.com/questions/14166/…. – Marijn Dec 2 '20 at 10:55
• I upvoted this answer since it can be useful for other people but in my case I do not want to migrate my whole document to minted, that's why I trying to fix it using listings. – Hugo Trentesaux Dec 2 '20 at 11:03


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Install the PARALLEL module

Parallel solver module for Chrono.

Features

The PARALLEL module provides features for performing multibody simulations using parallel computing within Chrono

• introduces a custom ChSystemParallel class
• implements a high-performance parallel collision detection algorithm
• uses efficient APIs for parallelism (OpenMP, Thrust, etc.)
• planned: support of MPI parallelism

Requirements

• To build this module you need:
The latest version of Blaze (3.4) does not work with Visual Studio. Use an earlier version (e.g. 3.0).
The easiest way to obtain the Thrust library is by installing the CUDA SDK. Alternatively, you can download or clone Thrust from its GitHub repository. In that case, you will need to manually specify the path to the Thrust headers during CMake configuration (set the variable THRUST_INCLUDE_DIR).

Building instructions

1. Download and install these libraries (depending on the platform, the process can be different)
2. Repeat the instructions for the full installation, but when you see the CMake window, you must add the following steps:
3. Set the ENABLE_MODULE_PARALLEL as 'on', then press 'Configure' (to refresh the variable list)
4. Set the path for BLAZE_DIR
5. Set the path for BOOST_DIR
1. Press 'Configure' again, then 'Generate', and proceed as usual in the installation instructions.
Not all the features of the standard serial version of Chrono::Engine are supported.

How to use it

• Look at the API section of this module for documentation about classes and functions.
• Look at the C++ source of demos to learn how to use the functions of this module.


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## A new fixed point theorem and applications.(English)Zbl 1272.54034

Summary: A new fixed point theorem is established under the setting of a generalized finitely continuous topological space (GFC-space) without the convexity structure. As applications, a weak KKM theorem and a minimax inequalities of Ky Fan type are also obtained under suitable conditions. Our results are different from known results in the literature.

### MSC:

 54H25 Fixed-point and coincidence theorems (topological aspects) 47H10 Fixed-point theorems
Full Text:

### References:

 [1] Tarafdar, E., A fixed point theorem and equilibrium point of an abstract economy, Journal of Mathematical Economics, 20, 2, 211-218, (1991) · Zbl 0718.90014 [2] Lan, K. Q.; Webb, J., New fixed point theorems for a family of mappings and applications to problems on sets with convex sections, Proceedings of the American Mathematical Society, 126, 4, 1127-1132, (1998) · Zbl 0891.46004 [3] Ansari, Q. H.; Yao, J.-C., A fixed point theorem and its applications to a system of variational inequalities, Bulletin of the Australian Mathematical Society, 59, 3, 433-442, (1999) · Zbl 0944.47037 [4] Ansari, Q. H.; Idzik, A.; Yao, J.-C., Coincidence and fixed point theorems with applications, Topological Methods in Nonlinear Analysis, 15, 1, 191-202, (2000) · Zbl 1029.54047 [5] Ding, X. P.; Tan, K.-K., Fixed point theorems and equilibria of noncompact generalized games, Fixed Point Theory and Applications, 80-96, (1992), Singapore: World Sicence, Singapore · Zbl 1404.47007 [6] Ding, X. P., Continuous selection, collectively fixed points and system of coincidence theorems in product topological spaces, Acta Mathematica Sinica, 22, 6, 1629-1638, (2006) · Zbl 1117.54032 [7] Lin, L.-J.; Ansari, Q. H., Collective fixed points and maximal elements with applications to abstract economies, Journal of Mathematical Analysis and Applications, 296, 2, 455-472, (2004) · Zbl 1051.54028 [8] Khanh, P. Q.; Long, V. S. T.; Quan, N. H., Continuous selections, collectively fixed points and weak Knaster-Kuratowski-Mazurkiewicz mappings in optimization, Journal of Optimization Theory and Applications, 151, 3, 552-572, (2011) · Zbl 1244.90221 [9] Ding, X. P.; Wang, L., Fixed points, minimax inequalities and equilibria of noncompact abstract economies in FC-spaces, Nonlinear Analysis A, 69, 2, 730-746, (2008) · Zbl 1157.47037 [10] Ding, X. P., Collective fixed points, generalized games and systems of generalized quasi-variational inclusion problems in topological spaces, Nonlinear Analysis A, 73, 6, 1834-1841, (2010) · Zbl 1229.54054 [11] Yuan, G. X.-Z., KKM Theory and Applications in Nonlinear Analysis, 218, (1999), New York, NY, USA: Marcel Dekker, New York, NY, USA [12] Balaj, M., Weakly $$G$$-KKM mappings $$, G$$-KKM property, and minimax inequalities, Journal of Mathematical Analysis and Applications, 294, 1, 237-245, (2004) · Zbl 1053.54028 [13] Balaj, M.; O’Regan, D., Weak-equilibrium problems in $$G$$-convex spaces, Rendiconti del Circolo Matematico di Palermo, 57, 1, 103-117, (2008) · Zbl 1225.54013 [14] Tang, G.-S.; Zhang, Q.-B.; Cheng, C.-Z.$$, W-G-F$$-KKM mapping, intersection theorems and minimax inequalities in FC-space, Journal of Mathematical Analysis and Applications, 334, 2, 1481-1491, (2007) · Zbl 1123.49002 [15] Hai, N. X.; Khanh, P. Q.; Quan, N. H., Some existence theorems in nonlinear analysis for mappings on GFC-spaces and applications, Nonlinear Analysis A, 71, 12, 6170-6181, (2009) · Zbl 1188.49006 [16] Khanh, P. Q.; Quan, N. H., Intersection theorems, coincidence theorems and maximal-element theorems in GFC-spaces, Optimization, 59, 1, 115-124, (2010) · Zbl 1185.49007 [17] Khanh, P. Q.; Quan, N. H., General existence theorems, alternative theorems and applications to minimax problems, Nonlinear Analysis, 72, 5, 2706-2715, (2010) · Zbl 1192.49015 [18] Park, S., Continuous selection theorems in generalized convex spaces, Numerical Functional Analysis and Optimization, 20, 5-6, 567-583, (1999) · Zbl 0931.54017 [19] Tan, K.-K., Comparison theorems on minimax inequalities, variational inequalities, and fixed point theorems, Journal of the London Mathematical Society, 28, 3, 555-562, (1983) · Zbl 0497.49010 [20] Park, S., Generalized Fan-Browder fixed point theorems and their applications, Collection of Papers Dedicated to J. G. Park, 51-77, (1989) [21] Liu, F.-C., On a form of KKM principle and Sup Inf Sup inequalities of von Neumann and of Ky Fan type, Journal of Mathematical Analysis and Applications, 155, 2, 420-436, (1991) · Zbl 0734.47030 [22] Kim, I., KKM theorem and minimax inequalities in $$G$$-convex spaces, Nonlinear Analysis Forum, 6, 1, 135-142, (2001) · Zbl 0993.54041
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Extension of Thurston's Geometrization Conjecture to four dimensions?

1. Apr 25, 2005

### straycat

Hello all,

I was wondering whether there might be anything similar to Thurston's Geometrization Conjecture, except for four manifolds instead of three manifolds. (Essentially, this conjecture (iiuc) states that it is always possible to take an arbitrary 3-manifold and break it up into smaller "pieces" of which there are eight types, each of which is pretty well understood -- see the explanation in mathworld [1].) I think that in two dimensions it is called the "uniformization theorem" [2], and ref [2] states that "It would be very important if similar results could be proved in higher dimensions." So I was wondering whether anyone has even *stated* the idea in dim=4?

David

[1] http://mathworld.wolfram.com/ThurstonsGeometrizationConjecture.html

[2] http://pdg.cecm.sfu.ca/~warp/java/uniform/node2.html [Broken]

Last edited by a moderator: May 2, 2017
2. Apr 25, 2005

### mathwonk

the discussions cited are not very precise, since they just use the word "geometry" without further explanation, but it appears from the analogy with 2 dimensions, that the meaning is something like riemannian geometry, i.e. metrics and curvature are involved.

in dimension 2 a compact connected manifold can be given a metric of constant curvature, and that curvature is zero iff the manifold is a torus, and positive iff the manidfold is a sphere and negative iff the manifold is of genus at least 2.

so apparently thurston is discussing the metric geometry of three manifolds. I do not know if such conjectures exist in higher dimensions, but the poincare conjecture, one of the corollaries of thurstons work it seems, already were proved in all higher dimewnsions, so for some questions, the three dimensional case was actually the hardest and the last one settled.

Apparently this is not the case for metric questions. There are however other popints of view toward generalizing the 2 dimensional uniformization theorem, such the shafarevich conjecture studied recently by Kolla'r and then others. this asks what the complex geometry is for universal covering spaces of algebraic varieties, in particular when they are holomorphically convex.

the analogy is that the 2 dimensional uniformization theorem follows from the fact that manifolds of geneus zero are spheres hence have well known positive curvature, while compact manifolds of genus zero are covered naturally by the plane which is flat, and complex riemann surfaces of genus at least 2 are covered by the unit disc, or upper half plane, which has hyperbolic geometry, i.e. negative curvature. these metrics and curvatures then descend to the compact manifolds they cover.

hence understanding the universal cover sheds light on the manifold covered. in this regard higher dimensions are more difficult since the number of possible covering manifolds is (literally) infinitely greater.

Last edited: Apr 25, 2005
3. Apr 26, 2005

### Doodle Bob

The one thing (and probably the most important thing) that is left out, is what Thurston means by a "geometry". In his book, he lays them out as such:

A model geometry (G,X) is a manifold X with a Lie group G of diffeomorphisms of X such that
1. X is connected and simply connected
2. G acts transitively on X, with compact point stabilizers
3. G is not contained in any larger group of diffeomorphisms of X satisfying #2 above, and
4. there is at least one compact manifold modeled on (G,X), i.e. as noted by math wonk, this would be a compact manifold with X as a universal cover.

A pretty straightforward definition, but I do not know if anyone has tried finding all of these in 4-dimensions. I do know that all homogeneous complex surfaces were classified a few years back. So, that's a start... However, there are plenty of 4-manifolds out there with no global complex structure, e.g. 1-dimensional quaternionic projective space...

-DB


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# graphite specific heat

Download PDF. Phys. The following is the relationship of several materials and heat three important parameters (thermal conductivity, specific heat capacity and density) of the reference value: Material Thermal conductivity (W/mK) Specific heat capacity (J/kg.K) Density (g/cm3) Aluminum 200 880 2.7 What is the specific heat capacity of graphite if it requires 266 J to raise the temperature of 15 grams of graphite by 25°C? Tool-Box. What is the specific heat of graphite? The specific heat of each concrete is shown in Figure 3. Graphite has a specific heat of 0.709 J/(gx 0 C). The following is the relationship of several materials and heat three important parameters (thermal conductivity, specific heat capacity and density) of the reference value: Material Thermal conductivity (W/mK) Specific heat capacity (J/kg.K) Density (g/cm3) Aluminum 200 880 2.7 Specific heat means the heat to raise the temperature of 1 gram of the material by 1 degree C. So if you want to raise the temperature of that one gram from 19 to 35 degrees, that same gram will need 35-19=16 times as much heat. Fig. 1.4 J/g°C B. The thermal conductivity (K) is the time rate of transfer of heat by conduction. D. 0.60 J/g°C. The interatomic spacing between the carbon atoms of graphite is a function of temperature. HT Pyrolytic Graphite vs. CuMo heat spreaders for a laser diode heat flux of 200 W/cm . Graphite has a specific heat of 0.709 J/(g×°C). Measurement of heat capacity and thermal conductivity of HDPE/expanded graphite nanocomposites by differential scanning calorimetry. In the case of high-purity Acheson graphite, these results are very close to the present values. This paper reviews the measurements and evaluations of the specific heat at constant pressure for graphites in the temperature range 200 K to 3500 K. A polynomial fit to the data is made for use with nuclear graphite, valid over the range 250 K–3000 K. -4,200,000 J 200 g of water in a glass is at 10.0°C when an ice cube is placed in the glass. Like its forerunner, pyrolytic graphite, GRAFOIL flexible graphite is resistant to heat, has no water of crystallization, is naturally lubricious, is chemically inert, and is an excellent conductor of heat and electricity. Dean (ed), Lange's Handbook of Chemistry (15th Edition), McGraw-Hill, 1999; Section 6, Thermodynamic Properties; Table 6.4, Heats of Fusion, Vaporization, and Sublimation and Specific Heat at Various Temperatures of the Elements and Inorganic Compounds 789-802. As the horizontal heat conduction materials, both have high thermal conductivity, lightweight, flexible and easy to be cut. Thermal conductivity increased from 0.442 W/m.K for pure HDPE to 0.938 W/m.K for nanocomposites containing 7% by weight of expended graphite. The characteristic anomaly at the superconducting transition (Tc = 11.4 K) indicates clearly the bulk nature of the superconductivity. Abstract: The specific heat of para-hydrogen films has been measured from 1 to 20 K for areal densities between 0.01 and 0.055 atoms/Å 2 of the partial monolayer adsorbed on grafoil. SIGRATHERM ® products for intelligent thermal management . The specific heat of para-hydrogen films has been measured from 1 to 20 K for areal densities between 0.01 and 0.055 atoms/Å 2 of the partial monolayer adsorbed on grafoil. GMT – This grade of graphite is produced as tubes for fluxing and other applications. Copper has a specific heat of 0.385 J/(gx 0 C). E. none of these Specific Heat of Para-Hydrogen Monolayers on Graphite 481 graphite are unfortunately not available so the present analysis incorporates the helium film values of 108 cc STP monolayer capacity and 0. ! Pyrolytic graphite (PG) is a unique form of graphite manufactured by decomposition of hydrocarbon gas at very high temperature in a vacuum furnace. Graphite Tubes. Consequently, heat treatment of graphite is an emerging market and graphite has become a vital material for numerous applications worldwide. Unlike manufactured pyrolytic graphite, GRAFOIL material is flexible, compactible, conformable and resilient. 1, No. These specific heat of carbon are made from food-grade materials so that you can also use them for packing foods. This comparison is used to validate the temperature-dependent thermal conductivity and specific heat capacity used in the models of the graphite foam. 0.71 J/g°C. 3. Pokatilov Laboratory of Physics and Engineering of Nanomaterials, Department of Physics and Engineering, Moldova State University, Chisinau MD-2009, Republic of Moldova Calories 575 Calories of heat are added to a sample of graphite (Specific Heat = 0.170 ). b) The specific heat capacity of graphite is 0.71 J/°C g. Calculate the energy required to raise the temperature of 30 kg graphite from 294 K to 348 K. The above grades exhibit excellent properties like, good thermal conductivity, low electrical specific resistance, low co-efficient of thermal expansion, good chemical inertness and can be easily machined. Graphite for industrial use is a synthetic material created from crystalline carbon. q = amount of heat (J) m = mass (grams) c = specific heat (J/g°C) ΔT = change in temperature (°C) 2. The specific heat capacity of graphite is 0.71 J/°C g. Calculate the energy required to raise the temperature of 34 kg graphite from 294 K to 348 K. J Custom design for your specific application – optimizes your process. The specific heat capacity of graphite is 0.71 J/°C g. Calculate the energy required to raise the temperature of 2.4 mol of graphite by 25.0°C. Graphite crucibles are used in fuel fired, electric, and induction furnaces or as a method for transferring and moving molten metals. Y. Aydogdu. Thermodynamics: Heat and Enthalpy Material Specific Heat Capacity (J/gºC) C (graphite) 0.720 Fe 0.451 Cu 0.385 Au 0.128 The constants of the elastic force acting between neighboring plates are derived from the measured elastic constants of graphite (mainly the compressibility). The 'best' data for the graphite specific heat in the range 200 K-3500 K. Debye pattern; the second rise is due to the formation CJ3R ~ 1 + D'/T2 + E'/T3 + F'/T4 + B'T . Made of modular, stacked graphite blocks with mounted PTFE seals between each adjoining block, the blocks are drilled in a specific pattern […] Ismail Tavman. Pyrolitic graphite is derived from a carbonized polymer and calendared to extraordinarily thin sheets. THERMAL CONDUCTIVITY Phonon hydrodynamics and ultrahigh–room-temperature thermal conductivity in thin graphite Yo Machida 1*, Nayuta Matsumoto ,Takayuki Isono1, Kamran Behnia2* Allotropes of carbon, such as diamond and graphene, are among the best conductors of heat. STRUCTURE Thermodynamically, graphite at atmospheric pressure is the more stable form of carbon. Graphite has a specific heat capacity of 0.709Jg ∘C. For decades, Mersen has designed and manufactured a comprehensive portfolio of heat exchangers using various technologies (blocs, tubes) and the materials (graphite, SiC, tantalum, zirconium, titanium). HPMS Thermally Conductive Graphite offers thermal conductivity up to 1800 W/m-K and are very thin, which allows for easy attachment to a variety of heat sink or component surfaces. By heating graphite up to 3000 °C, its properties are enhanced. The specific heat of a two-layer $^{4}\mathrm{He}$ film adsorbed on a graphite substrate is estimated as a function of temperature by quantum Monte Carlo simulations. Consequently, heat treatment of graphite is an emerging market and graphite has become a vital material for numerous applications worldwide. They are mostly used as secondary condensers after larger primary condensers. The structure of graphite consists of a succession of layers parallel to the basal plane of hexagonally linked carbon atoms. Vertical GE41 graphite block heat exchanger with custom made expansion system and a demountable header. 4. 1.4 J/g°C. Heat Transfer Engineering and Design. Heat is not the same as temperature, yet … HPMS Thermally Conductive Graphite offers thermal conductivity up to 1800 W/m-K and are very thin, which allows for easy attachment to a variety of heat sink or component surfaces. Figure 1: Schematic diagram of specific heat of graphite experiment Heater Filament Graphite Sample To thermocouple V Variac A . mediha kok. According the specific thermal 1 Thermal diffusion diagram of TCGS-S graphite heat sink sheet . The high thermal conductivity graphite can easily be used for thermal management in electronics (adhesives available for easier applying) or other space-limiting areas. Calculate the energy required to raise the temperature of 1.0 mol of graphite by 1.0 oC. 1 Thermal diffusion diagram of TCGS-S graphite heat sink sheet . study the specific heat of graphite in the fifties of the last century [3, 6, 7]. heat transfer. J.A. Hydrodynamic heat transport. Under vacuum or inert gas, graphite is very temperature-resistant which makes it a potential material for high temperature applications. The specific heat of high‐purity Acheson graphite prepared by the National Carbon Company has been measured from 13° to 300°K. Fuel Fired Furnace The specific heat capacity of graphite is 0.71 J/°C g. Calculate the energy required to raise the temperature of 4.2 mol of graphite by 25.0°C. What is the specific heat capacity of graphite if it requires 266 J to raise the temperature of 15 grams of graphite by 25°C? The superconducting state of Ca-intercalated graphite CaC6 has been investigated by specific heat measurements. The results are compared with the specific heat of 4 He films at equivalent densities. Under vacuum or inert gas, graphite is very temperature-resistant which makes it a potential material for high temperature applications. The maximum expansion potential of a specific grade of intercalated graphite can only be realized if the heating rate used to affect exfoliation is rapid. GRAFOIL The specific heat and effective thermal conductivity of composites containing single-wall and ... bulk graphite powder. The specific heat capacity of a particular graphite sample is determined to be 706.9 J K−1 kg−1 with a combined relative standard uncertainty of 9 parts in 104 at 295.15 K. Polynomial fit to data for C- The Debye theory of specific heat leads to an ex- pression which has no exact analytic form. Worldwide heat exchanger leader with solution-oriented approach.  Specific heat : The quantity of heat required to raise the temperature of one gram of the substance by one degree Celsius ( 1 ° C ) , The unit used in measuring specific heat is J/g . The specific heat of para-hydrogen films has been measured from 1 to 20 K for areal densities between 0.01 and 0.055 atoms/Å 2 of the partial monolayer adsorbed on grafoil. This test will validate the TES performance in an scenario where it is used as combined heat exchanger and storage for the Na -> CO₂ configuration. Rev. The results are compared with the specific heat of 4 He films at equivalent densities. Aim:- To determine the specific heat of graphite at different temperatures Specific Heat: Value given for solid graphite form. Recent experiments have shown that the specific heat of graphite varies as T 2 instead of T 3 between 15°K and 80°K. Our SIGMA G-TUBE graphite tube heat exchangers are ideal for use with corrosive materials. 1 atoms/A 2 at monolayer completion.1 Results for the four … 2. In this study, the effect of graphite nanofibers (diameter = 2 to 1000 nm, length = 100μm) on the bulk thermal properties of paraffin PCM (T melt = 56 °C) is investigated. Carbon and graphite fiber-based materials ensure high-quality insulation, equipment integrity, consistent processing parameters, energy savings, and personal safety. Endothermic or exothermic?_____ 5. … The conductivity of some graphite composites are being measured as high as 1,000-2,000 W/m°K [11]. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … GraphiteInsulation.com is an e-commerce platform for engineers, maintenance managers, and procurement oﬃces to research, purchase and explore various materials employed inside graphite hot zones.. Our Story. • Unique mechanical strength: the tensile, compressive and flexural strength of graphite increases as temperature increases to 2700 K. Alpaslan Turgut. 1 Measurement of specific heat of graphite Aim: Estimation of the specific heat of graphite Apparatus: (i) Graphite sample, (ii) Heater filament, (iii) Stop watch (iv) Variac (variable transformer), (v) Voltmeter, (vi) Ammeter and (vii) Chromel (Nickel-Chromium alloy) Alumel A sample of mercury has a mass of 500 grams. Download PDF. So as an example, if you had a pot filled with 1 kg of water and wanted to heat its temperature by 1°C, it would require 4,184 Joules of heat. Graphite is recognised for being an excellent conductor of heat and electricity, for exhibiting the highest natural strength and stiffness over all other minerals, for resistance to chemical attack and for withstanding high temperatures (higher than 3600 °C (6500 °F)). The outcome is discussed and the experimental uncertainty is reviewed. In the region 13° to 54°K the C p data follows a T 2 dependence quite accurately in agreement with previous experimental work and recent theoretical investigations of specific heat in strongly anisotropic solids. A. }, abstractNote = {Fundamental measurements have been obtained in the INL Graphite Characterization Laboratory to deduce the temperature dependence of thermal conductivity for G-348 isotropic graphite, which has been used by … The density of graphite blocks can be adjusted by it's raw materials, formulation and manufacturing processes used to create the specific material grade during initial production. Graphite/PCM concrete shows a rapid increase in specific heat in a temperature range between 22 and 30°C. And if you want to raise the temperature of 193 grams in stead of just 1 gram, you will need 193 times as much heat. The specific heat capacity of a particular graphite sample is determined to be 706.9 J K −1 kg −1 with a combined relative standard uncertainty of 9 parts in 10 4 at 295.15 K. The specific heat capacity of cyanoacrylate has also been determined. Graphite Insulation Mission Statement. Alpaslan Turgut. Ismail Tavman. °C . PGS is a competitive conductive sheet with high thermal conductivity and high flexibility. PGS Graphite Sheet PGS (Pyrolytic Highly Oriented Graphite Sheet) is made of graphite with a structure that is close to a single crystal, which is achieved by the heat decomposition of polymeric film. These graphite heat exchangers offer a near universal corrosion resistance and are of the highest strength on the market. impregnated graphite and silicon carbide result in heat exchangers that are applied successfully in both, mono and multi purpose plants. For conventional concrete, the peak point of the specific heat was not shown in the graph, indicating that concrete does not express a latent heat storage characteristic. mediha kok. Materials with a high thermal conductivity (materials such as these) are often used in applications where heat energy needs to be dissipated. In heat transfer analysis, thermal diffusivity is the thermal conductivity divided by density and specific heat capacity at constant pressure.It measures the rate of transfer of heat of a material from the hot side to the cold side. How much heat is added to 125 g of iron (Specific Heat = 0.108 ) if its temperature rises from 72.4 o C to 98.1 o C? The patented k-Core ® heat transfer system uses encapsulated graphite to help alleviate heat in high-power electronics for applications in aerospace, military, and commercial applications. The specific heat of material can be calculated by the following equation: where C p and C PR are the specific heat of the test sample and reference ... Due to excellent thermal conductivity, heat flux of graphite is higher than that of matrix, which indicates that the heat transmission would pass through the graphite preferentially. Some materials have high specific heat capacities, such as water or hydrogen, and other materials have low specific heat capacities, such as glass, copper, and lead. C . The temperature and magnetic field dependence of the electronic specific heat are consistent with a fully gapped superconducting order parameter. Graphite enhances the absorption and emission properties of the sample, which usually allows for more accurate measurements. As temperature rises, electrical conductivity increases. Low thermal mass allows rapid heat up and cool down. If a 20 gram piece of graphite is cooled from 38⁰C to 18⁰C, how much energy was lost by the graphite? In other words, the mean interatomic spacing increases and the result is thermal expansion. You need to look up the specific heat values (c) for aluminum and water. The intensive properties c v and c p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p) , respectively: Calculate the energy required to raise the temperature of 750 g of graphite by 160°C. However, phonons dominate Thermal Diffusivity Table . Graphite/PCM concrete shows a rapid increase in specific heat in a temperature range between 22 and 30[degrees]C. Soft-felt insulations or rigidized insulation structures can be used to protect high-temperature environments and specialized equipment, such as this graphite-lined vacuum furnace (Centorr Industries). Wherever heat or cold has to be managed or stored under extreme conditions within the vehicle, SGL Carbon’s SIGRATHERM products made of flexible graphite can provide solutions – due to their high thermal conductivity and their fast response characteristic compared to standard heat management systems. Different substances have different heat capacities. Measurement of heat capacity and thermal conductivity of HDPE/expanded graphite nanocomposites by differential scanning calorimetry. The specific heat capacity of a particular graphite sample is determined to be 706.9 J K −1 kg −1 with a combined relative standard uncertainty of 9 parts in 10 4 at 295.15 K. The specific heat capacity of cyanoacrylate has also been determined. The specific heat capacity of graphite is 0.71 \\mathrm{~J} / \\mathrm{g}. (1956). The thermal properties of conductivity and expansion are strongly influenced by the anisotropy of the graphite crystal. C. 0.43 J/g°C. , Low-temperature specific heat of the graphite intercalation compounds KC 8, CsC 8, RbC 8, and their parent highly oriented pyrolytic graphite. When it absorbs 280 joules, the temperature is found to have increased from 25°C to 29°C. Specific heat of twisted bilayer graphene: Engineering phonons by atomic plane rotations Denis L. Nika,1,2 Alexandr I. Cocemasov,1 and Alexander A. Balandin2,a) 1E. 17– 19 The speciﬁ c heat is stored by the lattice vibrations (phonons) and the free conduction elec-trons of a material, C = C p + C e . The k-Core ® material uses APG as an insert within an encapsulating structure. Specific heat of Graphite is 720 J/g K. Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. Recent studies on low-temperature laser-heated graphite suggest a complex energy exchange when compared to other materials. The result is an ultra-pure product which is near theoretical density and extremely anisotropic. The Philosophical Magazine: A Journal of Theoretical Experimental and Applied Physics: Vol. Joules. C p,solid: Constant pressure heat capacity of solid: S° solid,1 bar Entropy of solid at standard conditions (1 bar) Thermodynamics: Heat and Enthalpy Material Specific Heat Capacity (J/gºC) C (graphite) 0.720 Fe 0.451 Cu 0.385 Au 0.128 What is the specific heat of ice? The major applications of The specific heat of each concrete is shown in Figure 3. All materials are capable of being easily CNC machined or water jet cut into application specific, intricate shapes for use in high temperature environments up to 3,000 degrees C. Figure 1. Q. Graphite has a specific heat of 0.709 J/g⁰C. Pyrolitic graphite is derived from a carbonized polymer and calendared to extraordinarily thin sheets. SIGRAFLEX high-purity flexible graphite foil and sheets made by SGL Carbon are smart solutions for heat treatment processes like hardening, carbonizing and brazing, for quartz glass and semiconductor production as well as for various other furnace applications. Figure 1: Schematic diagram of specific heat of graphite experiment Heater Filament Graphite Sample To thermocouple V Variac A . Natural graphite is an excellent conductor of heat and electricity, stable over a broad range of temperatures, and a highly refractory material with a high melting point of 3650 °C. For q= m c Δ T : identify each variables by name & the units associated with it. The specific heat of graphite is 0.71 J/g oC. Calculate the energy required to raise the temperature of 2.4 moles of graphite by… Graphite (/ ˈ ɡ r æ f aɪ t /), archaically referred to as plumbago, is a crystalline form of the element carbon with its atoms arranged in a hexagonal structure.It occurs naturally in this form and is the most stable form of carbon under standard conditions.Under high pressures and temperatures it converts to diamond.Graphite is used in pencils and lubricants. Download Full … Schmidt® SIGMA G-BLOC Graphite Block Heat Exchangers Our SIGMA G-BLOC graphite block heat exchangers are ideal for use with corrosive materials, offer nearly universal corrosion resistance, and are the highest strength on the market. So we can offer die cut graphite sheet, graphite tape and composite graphite sheet with adhesive on both sides. Heat … The test at the CSIRO Mayfield facility is designed to test the Graphite Energy TES in a supercritical CO₂ extraction loop at 200bar. It is made by high temperature heat treatment and has certain shape retention, self-supporting property, small specific gravity, low thermal conductivity, thermal shock resistance, ablation resistance and long service life. 4 PROPERTIES AND CHARACTERISTICS OF GRAPHITE POCO GRAPHITE, INC. If a 25 gram piece of graphite is cooled from 35 °C to 18 °C, how much energy was lost by the… For each simulation set, the density and thermal conductivity were varied according to Table 1 below. Small contributions of the electronic specific heat and the C p – C v correction are added to obtain the total specific heat. Ismail Tavman. In the region 13° to 54°K the C p data follows a T 2 dependence quite accurately in agreement with previous experimental work and recent theoretical investigations of specific heat in strongly anisotropic solids. Chemical: pigments, heat recovery on corrosive process, VCM, TDI, etc... 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This was a final project for a class in Machine Dynamics. The objective was to design a project that showcased all that we learned throughout the quarter. I decided to model a paddle-ball that utilized a strange sort of ball: a triangular ball. However, this would introduce some interesting behavior upon collision. The spring was also made discontinuous, so that only beyond a certain length would there be a spring constant, under this length, the spring acts like a string.

The behavior was modeled and solved using the Euler-Lagrange equations and Hamiltonian dynamics in Mathematica.

# Setting up the equations

A frame was assigned to each of the vertexes of the triangle and a set of transformation matrices was set up to convert the center-of-mass frame to the points on the vertex. The idea is that we want to detect collisions using the vertices, since the object was a triangle interacting with a line (our paddle). Due to numerical integration, we don't expect that two vertices would interact with the same line.

The constraint is the oscillating surface of the paddle, where the frequency is given by $f$: $\phi = \text{sin}[ft]-y[t]$

The kinetic energy of the system is given by the translational and rotational energy of the rigid body. The $V^b$ matrix of the body is determined by taking the inverse of the transformation matrix to the center of mass for the body multiplied by the time derivative of the transformation matrix, $g$:

$V^b = g^{-1} \dot{g}$

The potential energy is given by the sum of the gravitational potential energy of the ball and the potential energy stored in the discontinuous spring.

$k=\begin{cases} 0, & \sqrt{x^2+(y-\sin[ft])^2}< 2 \\ 5, & \sqrt{x^2+(y-\sin[ft])^2}\geq 2.\end{cases}$

Before collisions, the Euler Lagrange equations were set up as an unconstrained equations:\begin{align*} \frac{d}{dt}\frac{\delta L}{\delta \dot q} - \frac{\delta L}{\delta q} = 0 \end{align*}

and collisions are detected when the y position of any vertex is equal to the y-vertex of the paddle.

## Collision

Once we have the point of intersection, we want to resolve the incoming velocity and solve the collision. We want the ($x',y',\theta'$) of the center of mass after the collision, and we know the configuration when it comes in, so we obtain the Hamiltonian before and after the collision, which should be the same, and solve for the generalized momenta with the constraint being the ground (paddle). The number of unknowns that we have to solve for is 4: ($x,y,\theta,\lambda$), and since we have 4 equations, we should be able to find a solution.

## Picking the right solution

Mathematica will return a set of suitable solutions.  We want $\lambda \leq 0$ because a positive value would mean that the paddle sucked the ball in. Picking the right solution to the update laws would then allow ($x',y',\theta '$) to be obtained after impact and the Lagrangian equations can be solved with no constraints until the next impact is detected.

Solving in Mathematica has some limitations because of the numerical analysis method of NDsolve. If x or $\theta$ is set to zero initially, this will generate a complex infinity error. This is because the solution obtained from the Euler Lagrangian equations in free fall has a cot($\theta$) term in it multiplied by a very small factor. In reality this might be zero, but when $\theta = 0$, this will generate an infinity term, so I avoided this problem by giving a non-zero displacement to $x$ and $\theta$.