text:
# Kerberos

Kerberos is an authentication protocol originally developed by MIT and implemented by various vendors and systems; probably its best description is in the first paragraph that can be found in the MIT Kerberos site:

Kerberos is a network authentication protocol. It is designed to provide strong authentication for client/server applications by using secret-key cryptography. A free implementation of this protocol is available from the Massachusetts Institute of Technology. Kerberos is available in many commercial products as well.

In the WWW can be found hundreds of sites where the protocol is fully described but the MIT site remains the primarily place in which one can get lost getting details about Kerberos – given that version IV is a late-1980 definition, I wouldn’t say it’s a young boy.

Recently I got to support SSO registration to our proprietary platform through Kerberos from our legacy client, a Win32/.NET application written in C/C++/Managed C++/C#. At the beginning I was quite sure I would have found wrapper classes in .NET that would make the whole thing as easy as a pie… soon I had to face the cold, hard reality. For some kind of mysterious reason, Microsoft has decided that among the tons of classes that in .NET wrap native features, Kerberos doesn’t deserve a single (meaningful) class.

Briefly – why is it called “Kerberos”? Because just like the mythological three-headed dog at the entrance of the Hades, three actors are involved in the protocol: the Client, the Service Server and Authentication Server; usually client and server negotiates the protocol, setup the channel, prepares and shares the key and so on – with Kerberos the Authentication Server is the keys repository and guarantees that the whole thing can go on the right way.

In fact, it is a bit more complicated than this: for example, when we logon in an interactive session in Windows we are already kicking in the Kerberos thing – but let’s proceed in an orderly manner.

The first step is to get a Ticket Granting Ticket: this is released by the AS, together with a session key.
The second step is to get a Ticket Granting Service for the service with wich we want to interact – to do so we need the TGT.
Finally, we can use the TGS to setup the connection with the Service.

In Windows, Kerberos is implemented as part of the Security Support Provider layer, exposed through its interface (the SSPI): among the modules there is the LAN Manager or -of course- Kerberos; so, I didn’t expect to find functions such as “GetTicketGrantingTicket()” or “GetTicketGrantingService()” but at least something reasonably easy to use. It ended up that it is easier to find people that has decided to go through RFC4120 and re-implement the whole protocol than find pages that explains in a bunch of easy step how to ask to the AD a TGS to grant access to a AD-registered service. To me this seamed a totally crazy thing: Windows uses Kerberos (it’s the default authentication protocol used by Active Directory) so it’s obvious that it has an implementation of it, and since SPPI is a public interface there must be function calls I can use to leverage it.

## Getting the Ticket Granting Ticket

The first thing I (wrongly) thought I had to do was to get the TGT: “well, to get a TGS I have to pass by the TGT, so I probably have to get the TGT first”.

This starts with a connection to the Local Security Authority:

HANDLE mLsaHandle;
NTSTATUS Status;

Status = LsaConnectUntrusted(mLsaHandle);

all these functions return an NTSTATUS: I won’t use it in the short excerpts I’m writing here but it makes of course sense to check it; once we have a connection handle, we can use it to obtaint the Kerberos package ID:

ULONG mPackageId;
LSA_STRING Name;
Name.Buffer = (PCHAR)MICROSOFT_KERBEROS_NAME_A;
Name.Length = (USHORT)strlen(Name.Buffer);
Name.MaximumLength = Name.Length + 1;

Status = LsaLookupAuthenticationPackage(mLsaHandle, &Name, &mPackageId);

if everything went fine, we have now the ID with which we can ask for the TGT, using a properly formatted KERB_QUERY_TKT_CACHE_REQUEST message:

bool GetTicketGrantingTicket()
{

    NTSTATUS Status = 0;

    NTSTATUS SubStatus = 0;

KERB_QUERY_TKT_CACHE_REQUEST CacheRequest;
PKERB_RETRIEVE_TKT_RESPONSE pTicketResponse = NULL;
ULONG ResponseSize;

CacheRequest.MessageType = KerbRetrieveTicketMessage;

// Set LogonId to 0 to request the current user's logon session
CacheRequest.LogonId.LowPart = 0;
CacheRequest.LogonId.HighPart = 0;

Status = LsaCallAuthenticationPackage(
mLsaHandle,
mPackageId,
&CacheRequest,
sizeof(CacheRequest),
(PVOID*)&pTicketResponse,
&ResponseSize,
&SubStatus);

// check Status/SubStatus

// free the return buffer
LsaFreeReturnBuffer(pTicketResponse);

return true;
}

Great! Now that we have the TGT, getting the TGS will probably require some kind of similar sequence of steps.

## Getting the Ticket Granting Service

We have understood that getting tickets is a matter of querying the LSA with proper messages (after all, Kerberos is a message based protocol, right?); to get the TGS we perform another call using the “service name” for which we want the TGS:

PKERB_EXTERNAL_TICKET GetTicketGrantingService(const char* service)
{
NTSTATUS Status;
PKERB_RETRIEVE_TKT_REQUEST CacheRequest = NULL;
PKERB_RETRIEVE_TKT_RESPONSE CacheResponse = NULL;
PKERB_EXTERNAL_TICKET Ticket = NULL;
ULONG ResponseSize;
NTSTATUS SubStatus;
BOOLEAN Trusted = TRUE;
BOOLEAN Success = FALSE;
UNICODE_STRING Target = { 0 };
UNICODE_STRING Target2 = { 0 };
stringstream ss;

wchar_t * wService = GetWC(service); // converts service in wchar
InitUnicodeString(&Target2, wService); // properly init the UNICODE string

CacheRequest = (PKERB_RETRIEVE_TKT_REQUEST)LocalAlloc(LMEM_ZEROINIT, Target2.Length + sizeof(KERB_RETRIEVE_TKT_REQUEST));

CacheRequest->MessageType = KerbRetrieveEncodedTicketMessage;
CacheRequest->LogonId.LowPart = 0;
CacheRequest->LogonId.HighPart = 0;
CacheRequest->EncryptionType = KERB_ETYPE_RC4_HMAC_NT;
CacheRequest->CacheOptions = KERB_RETRIEVE_TICKET_AS_KERB_CRED;
Target.Buffer = (LPWSTR)(CacheRequest + 1);
Target.Length = Target2.Length;
Target.MaximumLength = Target2.MaximumLength;
CopyMemory(Target.Buffer, Target2.Buffer, Target2.Length);
CacheRequest->TargetName = Target;

delete wService;

Status = LsaCallAuthenticationPackage(
mLsaHandle,
mPackageId,
CacheRequest,
Target2.Length + sizeof(KERB_RETRIEVE_TKT_REQUEST),
(PVOID *)&CacheResponse,
&ResponseSize,
&SubStatus
);

// check errors

if (CacheRequest)
{
LocalFree(CacheRequest);
}

return Ticket;
}

finally, since we are good Windows citizens, we cleanup the session:

bool TearDown()
{
//...
// as usual, check errors
return true;
}

Great! Now we have the TGS and thus we can use it to open a session with the Service Server!

Wrong.

## Security contexts

The factual thing about the previous steps is that we really got a TGT and a TGS: we can check them and we’ll recognize some properties (the domain, for example); one thing that is strange to me is that to get a TGS we didn’t have to pass by the TGT – oh well, probably in a Windows session it is implicitly the TGT I already got when I started the interactive session.

The bad thing is that in a real scenario a simple TGS is not enough to establish a session with a service: the ticket used in the session startup contains some other properties that are not part of a TGS.

To get a ticket valid to start a client/service session we have to obtain a more detailed opaque token that the service will submit to the local security implementation to verify that the client has the permission to use the service. To do so we have first (as often happens in Windows) acquire a session handle and then initialise a new security context:

// get a new credentials handle
SecHandle credHandle;
SECURITY_INTEGER pts;
SECURITY_STATUS ss = AcquireCredentialsHandle(
NULL,
MICROSOFT_KERBEROS_NAME_W,
SECPKG_CRED_OUTBOUND,
NULL,
NULL,
NULL,
NULL,
&credHandle,
&pts);
// check errors...

then we can ask an opaque TGS:

wchar_t* spn = GetWC(servicePrincipalName);  // SPN in wide char

SecBuffer outSecBuffer;
SecBufferDesc output;

outSecBuffer.pvBuffer = NULL;
outSecBuffer.BufferType = SECBUFFER_TOKEN;
outSecBuffer.cbBuffer = 0;
output.ulVersion = SECBUFFER_VERSION;
output.cBuffers = 1;
output.pBuffers = &outSecBuffer;

CtxtHandle contextHandle;
ULONG pfContextAttr;
TimeStamp timeStamp;

SECURITY_STATUS secCtxRes = InitializeSecurityContext(
&credHandle,
NULL,
spn,
ISC_REQ_CONNECTION | ISC_REQ_ALLOCATE_MEMORY,
0,
SECURITY_NATIVE_DREP,
NULL,
0,
&contextHandle,
&output,
&pfContextAttr,
&timeStamp);

// check error

int count = 0;
while (secCtxRes == SEC_I_CONTINUE_NEEDED)
{ // this part is not tested
secCtxRes = InitializeSecurityContext(
&credHandle,
&contextHandle,
spn,
ISC_REQ_CONNECTION | ISC_REQ_ALLOCATE_MEMORY,
0,
SECURITY_NATIVE_DREP,
&output,
0,
&contextHandle,
&output,
&pfContextAttr,
&timeStamp);
if (count > 5) break;
}

output.pBuffers = NULL;
delete spn;

FreeContextBuffer(outSecBuffer.pvBuffer);
DeleteSecurityContext(&contextHandle);

// outSecBuffer.pvBuffer

In outSecBuffer.pvBuffer there’s finally our valid opaque ticket.

## Conclusion

Mastering Kerberos is not easy; quite often, from a developer perspective, we stumble upon these topics and at a first sight we think that such an operation (to be granted to use a service) should be a straightforward operation because we know that many times we need to save time for other tasks.

Our job here is not finished: we have to wrap the Win32 implementation with a proper .NET implementation (or start with C++/CLI right from the beginning), but this should be an easy exercise.


text:
Sum the Vertex Connections

Let's say you have a positive integer N. First, build a regular polygon, that has N vertices, with the distance between neighbouring vertices being 1. Then connect lines from every vertex, to every other vertex. Lastly, calculate the length of all lines summed up together.

Example

Given the input N = 6, build a hexagon with lines connecting every vertex with the other vertices.

As you can see, there are a total of 6 border lines (length=1), 3 lines that have double the border length (length=2) and 6 other lines that we, by using the Pythagoras Theorem, can calculate the length for, which is

If we add the lengths of the lines together we get (6 * 1) + (3 * 2) + (6 * 1.732) = 22.392.

As structures with 2 or less vertices are not being considered polygons, output 0 (or NaN, since distance between a single vertex doesn't make much sense) for N = 1, since a single vertice cannot be connected to other vertices, and 1 for N = 2, since two vertices are connected by a single line.

Input

An integer N, in any reasonable format.

Output

The length of all the lines summed up together, accurate to at least 3 decimal places, either as a function return or directly printed to stdout.

Rules

• Standard loopholes are forbidden.
• This is , so the shortest code in bytes, in any language, wins.

Test Cases

(Input) -> (Output)
1 -> 0 or NaN
2 -> 1
3 -> 3
5 -> 13.091
6 -> 22.392

• Must we really handle 1? My current entry would return nan rather than zero for example, and would just require special casing for it. – Jonathan Allan Oct 23 '17 at 19:51
• @JonathanAllan I thought about it after seeing your answer, nan is fine too, as distance between a single vertex doesn't make much sense anyways. – Ian H. Oct 23 '17 at 19:54
• You should probably allow errors to be thrown too for n=1 I think. – Jonathan Allan Oct 23 '17 at 20:03
• It's hard to tell what 3 decimal places of accuracy means without an upper bound for N, since outputs get larger and floats get less precise. – xnor Oct 23 '17 at 20:16
• @xnor As long as it is precise up to 3 decimal places for any reasonable input N, its fine is the result is less precise for huge numbers. – Ian H. Oct 23 '17 at 20:26

Python 3(with sympy),  61 60 58 54  48 bytes

-6 (maybe even -10 if we do not need to handle n=1) thanks to xnor (further trigonometric simplification plus further golfing to handle edge case of 1 and save parentheses by moving a (now unnecessary) float cast).

Hopefully beatable with no 3rd party libraries? Yes!! but Lets get things rolling...

lambda n:1%n*n/2/(1-cos(pi/n))
from math import*

Try it online!

This uses a formula for the sum of the lengths if a polygon is inscribed inside a unit circle, n*cot(pi/2/n)/2 and adjusts the result to one for the side length being one by dividing by the sin of that cord length sin(pi/n).

The first formula is acquired by considering the n-1 cord lengths of all the diagonals emanating from one corner which are of lengths sin(pi/n) (again), sin(2*pi/n), ..., sin((n-1)pi/n). The sum of this is cot(pi/2/n), there are n corners so we multiply by n, but then we've double counted all the cords, so we divide by two.

The resulting n*cot(pi/2/n)/2/sin(pi/n) was then simplified by xnor to n/2/(1-cos(pi/n)) (holding for n>1)

...this (so long as the accuracy is acceptable) now no longer requires sympy over the built-in math module (math.pi=3.141592653589793).

• yes! saved 11 bytes. cool formula! – J42161217 Oct 23 '17 at 19:57
• It looks like the formula simplifies to n/2/(1-cos(pi/n)). – xnor Oct 23 '17 at 20:11
• Good spot @xnor (so long as we may output 0.25 for n=1 - but special casing may be shorter too...) – Jonathan Allan Oct 23 '17 at 20:17
• @JonathanAllan Huh, weird that 1/4 is the result for n=1. It can be patched with 1%n*. Also, parens can be saved by moving the float inside to float(1-cos(pi/n)). I don't know sympy much, but maybe there's an arithmetic way to force a float. – xnor Oct 23 '17 at 20:24
• @xnor Thanks! (I should have noticed the float move). sympy outputs an expression - e.g. for n=6 no cast results in an expression with a representation 3.0/(-sqrt(3)/2 + 1) - there may well be a shorter way but I don't know it yet. – Jonathan Allan Oct 23 '17 at 20:34

Python, 34 bytes

lambda n:1%n*n/abs(1-1j**(2/n))**2

Try it online!

Uses the formula n/2/(1-cos(pi/n)) simplified from Jonathan Allan. Neil saved 10 bytes by noting that Python can compute roots of unity as fractional powers of 1j.

Python without imports doesn't have built-in trigonometric functions, pi, or e. To make n=1 give 0 rather than 0.25, we prepend 1%n*.

A longer version using only natural-number powers:

lambda n:1%n*n/abs(1-(1+1e-8j/n)**314159265)**2

Try it online!

• Cool as a cucumber. – Jonathan Allan Oct 23 '17 at 20:57
• 37 bytes: lambda n:1%n*n/(1-(1j**(2/n)).real)/2 – Neil Oct 23 '17 at 21:24
• @Neil Wow, Python can just compute roots of unity. – xnor Oct 23 '17 at 21:29
• Well, that was the easy bit. I don't know what abs() does though. – Neil Oct 23 '17 at 21:43
• @Neil it gets the absolute value, hence the norm, i.e. the distance from the origin. – Jonathan Allan Oct 23 '17 at 21:44

MATL, 16 15 bytes

t:=ZF&-|Rst2)/s

This uses a commit which introduced the FFT (Fast Fourier Transform) function, and which predates the challenge by 8 days.

Explanation

The code uses this trick (adapted to MATL) to generate the roots of unity. These give the positions of the vertices as complex numbers, except that the distance between consecutive vertices is not normalized to 1. To solve that, after computing all pairwise distances, the program divides them by the distance between consecutive vertices.

t       % Implicit input, n. Duplicate
:       % Range: [1 2 ... n-1 n]
=       % Isequal, element-wise. Gives [0 0 ... 0 1]
ZF      % FFT. Gives the n complex n-th roots of unity
&-|     % Matrix of pairwise absolute differences
R       % Upper triangular matrix. This avoids counting each line twice.
s       % Sum of each column. The second entry gives the distance between
% consecutive vertices
t2)/    % Divide all entries by the second entry
s       % Sum. Implicit display

• this is beautiful – Jonah Oct 26 '17 at 10:47
• @Jonah Complex numbers FTW :-) – Luis Mendo Oct 26 '17 at 11:17

Grasshopper, 25 primitives (11 components, 14 wires)

I read a meta post about programs in GH and LabVIEW, and follow similar instructions to measure a visual language.

Print <null> for N = 0, 1, 2,because Polygon Primitive cannot generate a polygon with 2 or fewer edges and you will get an empty list of lines.

Components from left to right:

• Side count slider: input
• Polygon Primitive: draw a polygon on canvas
• Explode: Explode a polyline into segements and vertices
• Cross reference: build holistic cross reference between all vertices
• Line: draw a line between all pairs
• Delete Duplicate Lines
• Length of curve
• (upper) Sum
• (lower) Division: because Polygon Primitive draws polygon based on radius, we need to scale the shape
• Multipication
• Panel: output

Mathematica, 26 bytes

uses @Jonathan Allan's formula

N@Cot[Pi/2/#]/2Csc[Pi/#]#&

-1 byte junghwan min

• -1 byte: N@Cot[Pi/2/#]/2Csc[Pi/#]#& since 1/sin(x) = csc(x) – JungHwan Min Oct 23 '17 at 21:02
• .5Csc[x=Pi/#]Cot[x/2]#& – Misha Lavrov Oct 24 '17 at 0:50

f 1=0
f n=n/2/(1-cos(pi/n))

Try it online!

I just dove into Haskell, so this turns out to be a fair beginner golf (that is, copying the formula from other answers).

I've also tried hard to put \$ somewhere but the compiler keeps yelling at me, so this is the best I've got. :P

Jelly, 1312 11 bytes

Uses Jonathan Allan's formula (and thanks to him for saving 2 bytes)

ØP÷ÆẠCḤɓ’ȧ÷

Try it online!

I've always been pretty fascinated with Jelly, but haven't used it much, so this might not be the simplest form.

• Save a byte by using "argument swapping dyadic chain separation", ɓ, to inline your helper link like so: ØP÷ÆẠCḤɓn1×÷ – Jonathan Allan Oct 24 '17 at 19:45
• @JonathanAllan oh thanks, I'm still a beginner and knew there probably was a better way than having a new chain but didn't know how to do it – Jeffmagma Oct 24 '17 at 20:48
• Oh, we can save another by using decrement, ’, and logical-and, ȧ: ØP÷ÆẠCḤɓ’ȧ÷ :) – Jonathan Allan Oct 24 '17 at 20:59
• oh wow thanks I hadn't thought of that – Jeffmagma Oct 24 '17 at 21:07

Javascript (ES6), 36 bytes

n=>1%n*n/2/(1-Math.cos(Math.PI/n))

Port of @JonathanAllan's Python 3 answer

f=n=>1%n*n/2/(1-Math.cos(Math.PI/n))
<input id=i type=number oninput="o.innerText=f(i.value)" /><pre id=o>

format: plain


text:
# Dipole-Dipole Interactions

Dipole-Dipole interactions result when two dipolar molecules interact with each other through space. When this occurs, the partially negative portion of one of the polar molecules is attracted to the partially positive portion of the second polar molecule. This type of interaction between molecules accounts for many physically and biologically significant phenomena such as the elevated boiling point of water.

### Definition of a Dipole

Molecular dipoles occur due to the unequal sharing of electrons between atoms in a molecule. Those atoms that are more electronegative pull the bonded electrons closer to themselves. The buildup of electron density around an atom or discreet region of a molecule can result in a molecular dipole in which one side of the molecule possesses a partially negative charge and the other side a partially positive charge. Molecules with dipoles that are not canceled by their molecular geometry are said to be polar.

Example 1: Carbon Dioxide and Hydrogen Flouride

[Figure 1]                                                                                [Figure 2]

In figure 1 above, the more electronegative Oxygen atoms pull electron density towards themselves as demonstrated by the arrows. Carbon Dioxide is not polar however because of its linear geometry. A molecule's overall dipole is directional, and is given by the vector sum of the dipoles between the atoms. If we imagined the Carbon Dioxide molecule centered at 0 in the XY coordinate plane, the molecule's overall dipole would be given by the following equation:

$\mu \cos(0) + -\mu \cos(0) = 0.$

Where μ is the dipole moment of the bond (given by μ=Q x r where Q is the charge and r is the distance of separation). Therefore, the two dipoles cancel each other out to yield a molecule with no net dipole.

In contrast, figure 2 demonstrates a situation where a molecular dipole does result. There is no opposing dipole moment to cancel out the one that is shown above. If we were to imagine the hydrogen flouride molecule placed so that the Hydrogen sat at the origin in the XY coordinate plane, the dipole would be given by $$\mu \cos(0)=\mu$$.

### Potential Energy of Dipole Interaction

Potential energy is the maximum energy that is available for an object to do work.  In physics, work is a quantity that describes the energy expended as a force operates over a distance. Potential energy is positional because it depends on the forces acting on an object at its position in space. For instance, we could say that an object held above the ground has a potential energy equal to its mass x acceleration due to gravity x its height above the ground (i.e., $$mgh$$). This potential energy that an object has as a result of its position can be used to do work. For instance we could use a pulley system with a large weight held above the ground to hoist a smaller weight into the air. As we drop the large weight it converts its potential energy to kinetic energy and does work on the rope which lifts the smaller weight into the air. It is important to remember that due to the second law of thermodynamics, the amount of work done by an object can never exceed (and is often considerably less) than the objects potential energy.

On a subatomic level, charged atoms have an electric potential which allows them to interact with each other. Electric potential refers to the energy held by a charged particle as a result of it's position relative to a second charged particle. Electric potential depends on charge polarity, charge strength and distance. Molecules with the same charge will repel each other as they come closer together while molecules with opposite charges will attract.

For two positively charged particles interacting at a distance r, the potential energy possessed by the system can be defined using Coulomb's Law:

$$V = \dfrac{kQq}{r}$$

where

• $$k$$ is the Coulomb constant and
• $$Q$$ and $$q$$ refer to the magnitude of the charge for each particle in Coulombs.

The above equation can also be used to calculate the distance between two charged particles (r) if we know the potential energy of the system. While Coulomb's law is important, it only gives the potential energy between two point particles. Since molecules are much larger than point particles and have charge concentrated over a larger area, we have to come up with a new equation.

The potential energy possessed by two polar atoms interacting with each other depends on the dipole moment, μ, of each molecule, the distance apart, r, and the orientation in which the two molecules interact. For the case in which the partially positive area of one molecule interacts only with the partially negative area of the other molecule, the potential energy is given by:

$V = -\dfrac{2\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}}$

where ε is the permeability of space. If it is not the case that the molecular dipoles interact in this straight end to end manor, we have to account mathematically for the change in potential energy due to the angle between the dipoles. We can add an angular term to the above equation to account for this new parameter of the system:

$V=-\dfrac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r_{12}^{3}}{(\cos\theta_{12}- 3\cos\theta_{1}\cos\theta_{2})}$

In this formula $$\theta_{12}$$ is the angle made by the two oppositely charged dipoles, and $$r_{12}$$ is the distance between the two molecules. Also, $$\theta_{1}$$ and $$\theta_{2}$$ are the angles formed by the two dipoles with respect to the line connecting their centers.

It is also important to find the potential energy of the dipole moment for more than two interacting molecules. An important concept to keep in mind when dealing with multiple charged molecules interacting is that like charges repel and opposite charges attract. So for a system in which three charged molecules (2 positively charged molecules and 1 negatively charged molecule) are interacting, we need to consider the angle between the attractive and repellant forces.

The potential energy for the dipole interaction between multiple charged molecules is:

$$V = \dfrac{kp \cos\theta}{r^{2}}$$

where

• k is the Coulomb constant, and
• r is the distance between the molecules.

### Dipole-Dipole Interactions in Macroscopic Systems

It would seem, based on the above discussion, that in a system composed of a large number of dipolar molecules randomly interacting with one another, V should go to zero because the molecules adopt all possible orientations. Thus the negative potential energy of two molecular dipoles participating in a favorable interaction would be cancelled out by the positive energy of two molecular dipoles participating in a high potential energy interaction. Contrary to our assumption, in bulk systems, it is more probable for dipolar molecules to interact in such a way as to minimize their potential energy (i.e., dipoles form less energetic, more probable configurations in accordance with the Boltzmann's Distribution). For instance, the partially positive area of a molecular dipole being held next to the partially positive area of a second molecular dipole is a high potential energy configuration and few molecules in the system will have sufficient energy to adopt it at room temperature. Generally, the higher potential energy configurations are only able to be populated at elevated temperatures. Therefore, the interactions of dipoles in a bulk solution are not random, and instead adopt more probable, lower energy configurations. The following equation takes this into account:

$$V=-\dfrac{2\mu_{A}^2\mu_{B}^2}{3(4\pi\epsilon_{0})^2r^6}\dfrac{1}{k_{B}T}$$

### Biological Importance of Dipole Interactions

The potential energy from dipole interactions is important for living organisms. The biggest impact dipole interactions have on living organisms is seen with protein folding. Every process of protein formation, from the binding of individual amino acids to secondary structures to tertiary structures and even the formation of quaternary structures is dependent on dipole-dipole interactions.

A prime example of quaternary dipole interaction that is vital to human health is the formation of erythrocytes. Erythrocytes, commonly known as red blood cells, are comprised of four protein subunits and a heme molecule. For an erythrocyte to form properly, multiple steps must occur, all of which involve dipole interactions. The four protein subunits—two alpha chains, two beta chains—and the heme group, interact with each other through a series of dipole-dipole interactions which allow the erythrocyte to take its final shape. Any mutation that destroys these dipole-dipole interactions prevents the erythrocyte from forming properly, and impairs their ability to carry oxygen to the tissues of the body. So we can see that without the dipole-dipole interactions, proteins would not be able to fold properly and all life as we know it would cease to exist.

Example 1

Calculate the potential energy of the dipole-dipole interaction between 2 HF molecules oriented along the x axis in an XY coordinate plane whose area of positive charge is separated by 5.00 Angstroms from the area of negative charge of the adjacent molecule:

SOLUTION

The Dipole moment of the HF molecules can be found in many tables, μ=1.92 D. Assume the molecules exist in a vaccum such that $$\epsilon_{0}=8.8541878\times10^{-12}C^2N^{-1}m^{-2}$$

$\mu=1.92D\times3.3356\times10^{-30}\dfrac{Cm}{D}=6.4044\times10^{-30}Cm$

$V=-\dfrac{2(6.4044\times10^{-30}Cm)^2}{4(8.8541878\times10^{-12}C^2N^{-1}m^{-2})(5.0\times10^{-10})^3}=1.4745\times10^{-19}Nm$

$V=-1.47\times10^{-19}J$

Example 2

Now imagine the same two HF molecules in the following orientation:

Given: $$\theta_{1}=\dfrac{3\pi}{4}$$, $$\theta_{2}=\dfrac{\pi}{3}$$ and $$\theta_{12}=\dfrac{5\pi}{12}$$

SOLUTION

$V=-\dfrac{(6.4044\times10^{-30}Cm)^2}{4\pi(8.8541878\times10^{-12}C^2N^{-1}m^{-2}(5.00\times10^{-10}m)^3}(\cos\dfrac{5\pi}{12}-3\cos\dfrac{3\pi}{4}\cos\dfrac{\pi}{3})$

$$V=-9.73\times10^{-20}Nm=9.73\times10^{-20}J$$

Example 3

Using the above formula, calculate the average enery of HF molecules interacting with one another in a bulk solution assuming that the molecules are 4.00 Angstroms apart in room temperature solution.

SOLUTION

$V=-\dfrac{2}{3}\dfrac{(6.4044\times10^{-30}Cm)^4}{(4\pi(8.8541878\times10^{-12}C^2N^{-1}m^{-2})^2(4.00\times10^{-10}m)^6}\dfrac{1}{(1.381\times10^{-23}Jk^{-1})(298k)}$

$V=-5.46\times10^{-21}J$

Example 4
What is the amount of energy stabilization that is provided to the system when 1 mole of HF atoms interact through dipole-dipole interactions.

SOLUTION

Since we have already calculated above the average potential energy of the HF dipole-dipole interaction this problem can be easily solved.

$V=-5.46\times10^{-21}J\times(6.022\times10^{23}mol^{-1})$

$V=-3288\dfrac{J}{mol}=3.29\dfrac{kJ}{mol}$

Example 5

Looking at the above equation for the average potential energy of the dipole-dipole interaction, what happens to the potential energy of the interaction as temperature increases.

SOLUTION

The potential energy of the dipole-dipole interaction decreases as T increases. This can be seen from the form of the above equation, but an explanation for this observation is relatively simple to come by. As the temperature of the system increases, more molecules have sufficient energy to occupy the less favorable configurations. The higher, less favorable, configurations are those that give less favorable interactions between the dipoles (i.e., higher potential energy configurations).

### References

1. Le Fèvre, R. J. W. (1953). Dipole moments; their measurement and application in chemistry. London,, Methuen.
2. Atkins, P. W. and J. De Paula (2006). Physical chemistry for the life sciences. New York, Oxford University Press ; Freeman.
3. Petrucci, R. H., W. S. Harwood, et al. (2002). General chemistry : principles and modern applications. Upper Saddle River, N.J., Prentice Hall.
4. Bloomfield, M. M. (1992). Chemistry and the living organism. New York, Wiley.
5. Silbey, R. J., R. A. Alberty, et al. (2005). Physical chemistry. Hoboken, NJ, Wiley.
6. HyperPhysics. http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Georgia State University, Department of Physics and Astronomy. 2008.
7. Campbell, N. A. and J. B. Reece (2005). Biology. San Francisco, Pearson Benjamin Cummings.
8. Chang, Raymond. (2005). Physical Chemistry for the Biosciences. Sausalito, California, University Science Books.

### Contributors

• David Johns (UCD), Joel Guess (UCD)

09:55, 9 May 2015

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text:
anonymous one year ago check my answer? Find S_11 for 1 + 2 + 4 + 8 + ... I got 2047, is that correct?

1. anonymous

wht is "S"?

2. anonymous

I believe that it means the sum of the series

3. anonymous

Write out your equation for me.

5. anonymous

well I know that r=2 and so I multiplied by 2 for each and got 1,2,4,8,16,32,64,128,256,512,1024 and then I looked it up because I wasn't sure what to do next and they got 2047 and I wanted to know how they got that

There is an equation. First you have to find the t11. T11 = term 11.

7. anonymous

The eleventh term is 1024 I believe

8. anonymous

its geometric progression, a, ar, ar^2, ar^3.... the common ratio here is 2, that is, every next term is "previous term multiplied by 2" sum in a GP is given by the formula $\frac{ a (1-r ^{n})}{ 1-r }$

9. anonymous

take a as 1, r as 2 and n as 11

10. anonymous

(1(1-2^11))/1-2 =1-2048/1-2 =2047


text:
Synopsis

# Freeze frame

Physics 4, s22
Images reveal some of the factors that determine the shapes of icicles.

At a glance, the icicles dangling from the edge of a house on a cold day all look pretty much like long, inverted cones. Look more closely though, and it is possible to see they are not perfectly conical, but more subtly curved, like a carrot.

Despite the many parameters involved, theorists have predicted that icicles, as well as other natural features like stalactites, should all converge to the same shape as they grow. In a paper appearing in Physical Review E, Antony Chen and Stephen Morris at the University of Toronto, Canada, describe an experimental setup that allows them to image icicles as they grow under controlled conditions, and test these predictions. They mounted a camera through a slot in the side of a refrigerator, within which icicles formed as water dripped from a nozzle and onto a rotating wooden support. Rotating the support helps even out the effects of drafts and temperature gradients.

Chen and Morris imaged $93$ icicles, documenting the shapes of fully grown icicles formed from both frozen tap and distilled water. For the most part, the shape of icicles follows a simple theoretical prediction: the diameter varies as the distance from the tip to the $3/4$th power (i.e., it is sublinear). Where this relation fails, one culprit is the purity of the water. Tap water has more impurities and a lower surface tension, which Chen and Morris find makes icicles more susceptible to bending, twisting, and forming ripples along their sides. – Jessica Thomas

## Subject Areas

Interdisciplinary Physics

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text:
# Harry R. Schwartz

Software engineer, nominal scientist, gentleman of the internet.
Member, ←Hotline Webring→.

· 1B41 8F2C 23DE DD9C 807E A74F 841B 3DAE 25AE 721B

ovo-lacto vegetarian

he

him

his

### Mamihlapinatapai

Published .
Tags: language, words.

Mamihlapinatapai is a word from the Yaghan language of Tierra del Fuego, listed in The Guinness Book of World Records as the “most succinct word,” and is considered one of the hardest words to translate. It describes a look shared by two people with each wishing that the other will initiate something that both desire but which neither one wants to start. This could perhaps be translated more succinctly as “eye-contact implying ‘after you…’” A more literal approximation is “ending up mutually at a loss as to what to do about each other.”

What a shame that English doesn’t have a word for this!

You might like these related articles:


text:
# Biot-Savart Law

A current element of length $\mathrm{d}\vec{l}$ carrying current $I$ produces a magnetic field at point P, \begin{align} \mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\;\mathrm{d}\vec{l}\times\hat{r}}{r^2}, \end{align} where $\vec{r}$ is the position vector of P from the current element.


text:
# Where was/is Compensated Compactness used?

This last summer, I read up on Tartar's so called Method of Compensated Compactness (or at least how it applied to scalar conservation laws). I used this theory to prove the existence of $L^{\infty}$ solutions to the scalar conservation law $u_{t}+(f(u))_{x}=0$ with initial data in $L^{\infty}$. Here we do not need to have a strictly hyperbolic conservation law. Perhaps just assume $f\in C^{2}$. The sources I used for my presentation were from Yunguang Lu's book Hyperbolic Conservation Laws and the Compensated Compactness Method and Dafermos's book Hypberbolic Conservation Laws in Continuum Physics.

I am aware that Compensated Compactness can be used to prove existence of solutions to $2\times 2$ systems of conservation laws. I haven't looked over this proof, but I believe Lu covers this.

My question is as follows: Where else is the method of Compensated Compactness used? If you can say, if possible, could you give a rough, brief sketch of how the method was used? Any input/thoughts would be appreciated.

-

Compensated compactness helps when one needs to find the limit of $u_n \cdot v_n$, where the sequences of vector fields $u_n$ and $v_n$ converge weakly in $L^2$: $u_n\rightharpoonup u$, $v_n\rightharpoonup v$. If none of the sequences converges strongly in $L^2$, the vector fields can still possess some additional properties which would compensate for the lack of strong convergence. For instance, the following result and its versions are often used in the homogenization theory.

Lemma. Let $u_n$, $v_n\in L^2(\Omega)$, where $\Omega$ is a bounded domain in $\mathbb R^d$ and let $u_n\rightharpoonup 0$, $v_n\rightharpoonup 0$ in $L^2(\Omega)$. Assume that $$\mbox{curl }v=0\qquad \mbox{and}\qquad \mbox{div } u_n\to w\quad \mbox{in }\ H^{-1}(\Omega).$$ Then $u_nv_n\to uv$ in the weak-* topology of $L^1(\Omega)$.

The lemma can be used to justify the convergence of solutions of the Dirichlet problem with strongly oscillating coefficients $$\mbox{div }\left(a\left(\frac{x}{\varepsilon}\right)\nabla u_\epsilon\right)=f,\qquad \left.u\right|_{\partial\Omega}=0,$$ to the solutions of some averaged problem (see, e.g., "Homogenization of differential operators and integral functionals" by V. V. Jikov, S. M. Kozlov, O. A. Oleinik).

-

In addition to proving the existence of solutions to a large class of nonlinear partial differential equations, among them conservation laws, compensated compactness is also used to show the convergence of numerical methods to approximate those solutions. Given an equation $P(u)=0$, one construcsts by whatever method (finite differences, finite elements, spectral,...) a sequence of approximations $u_n$. Then one tries to prove:

1. the sequence $\{u_n\}$ converges in a some sense to some function $u$;
2. this function $u$ is a solution of the original problem;
3. an eror estimate for $u-u_n$.

Then, if sufficiently good a priori estimates are known for $u_n$, the method of compensated compactness can give 1. and 2.This plan is carried out for instance here and here,

-


text:
# For a heat engine working on an air-standard Diesel cycle the efficiency will

This question was previously asked in
ESE Mechanical 2013 Official Paper - 1
View all UPSC IES Papers >
1. Increase as the fuel cut-off ratio is increased
2. Not change as the fuel cut-ratio is increased
3. Decrease as the fuel cut-off ratio is increased
4. Change as the fuel cut-off ratio is increased

## Answer (Detailed Solution Below)

Option 3 : Decrease as the fuel cut-off ratio is increased

## Detailed Solution

Concept:

The efficiency of standard diesel cycle is given as

$${\eta _{diesel}} = 1 - \frac{1}{{r_k^{\gamma - 1}}}\left( {\frac{{r_c^\gamma - 1}}{{{r_c} - 1}}} \right)$$

For $${r_c} = 1.65,\;\frac{{r_c^\gamma - 1}}{{{r_c} - 1}} = 1.563$$

For $${r_c} = 1.75,\frac{{\;r_c^\gamma - 1}}{{{r_c} - 1}} = 1.5853$$

So, $$\left( {\frac{{r_c^\gamma - 1}}{{{r_c} - 1}}} \right)$$ increases with increase in cut-off ratio.

∴ ηdiesel decreases


text:
# 7.4: Magnetic Units of Measurement

First, we need to become acquainted with the various quantities associated with magnetism. There are quite a few more quantities to be dealt with in magnetic systems than for electrical systems. With electricity, the basic quantities are Voltage (E), Current (I), Resistance (R), and Power (P). The first three are related to one another by Ohm’s Law (E=IR ; I=E/R ; R=E/I), while Power is related to voltage, current, and resistance by Joule’s Law (P=IE ; P=I2R ; P=E2/R).

With magnetism, we have the following quantities to deal with:

Magnetomotive Force—The quantity of magnetic field force, or “push.” Analogous to electric voltage (electromotive force).

Field Flux—The quantity of total field effect, or “substance” of the field. Analogous to electric current.

Field Intensity—The amount of field force (mmf) distributed over the length of the electromagnet. Sometimes referred to as Magnetizing Force.

Flux Density—The amount of magnetic field flux concentrated in a given area.

Reluctance—The opposition to magnetic field flux through a given volume of space or material. Analogous to electrical resistance.

Permeability—The specific measure of a material’s acceptance of magnetic flux, analogous to the specific resistance of a conductive material (ρ), except inverse (greater permeability means easier passage of magnetic flux, whereas greater specific resistance means more difficult passage of electric current).

But wait . . . the fun is just beginning! Not only do we have more quantities to keep track of with magnetism than with electricity, but we have several different systems of unit measurement for each of these quantities. As with common quantities of length, weight, volume, and temperature, we have both English and metric systems. However, there is actually more than one metric system of units, and multiple metric systems are used in magnetic field measurements! One is called the cgs, which stands for Centimeter-Gram-Second, denoting the root measures upon which the whole system is based. The other was originally known as the mks system, which stood for Meter-Kilogram-Second, which was later revised into another system, called rmks, standing for Rationalized Meter-Kilogram-Second. This ended up being adopted as an international standard and renamed SI (Systeme International).

And yes, the µ symbol is really the same as the metric prefix “micro.” I find this especially confusing, using the exact same alphabetical character to symbolize both a specific quantity and a general metric prefix!

As you might have guessed already, the relationship between field force, field flux, and reluctance is much the same as that between the electrical quantities of electromotive force (E), current (I), and resistance (R). This provides something akin to an Ohm’s Law for magnetic circuits:

And, given that permeability is inversely analogous to specific resistance, the equation for finding the reluctance of a magnetic material is very similar to that for finding the resistance of a conductor:

In either case, a longer piece of material provides a greater opposition, all other factors being equal. Also, a larger cross-sectional area makes for less opposition, all other factors being equal.

The major caveat here is that the reluctance of a material to magnetic flux actually changes with the concentration of flux going through it. This makes the “Ohm’s Law” for magnetic circuits nonlinear and far more difficult to work with than the electrical version of Ohm’s Law. It would be analogous to having a resistor that changed resistance as the current through it varied (a circuit composed of varistors instead of resistors).

This page titled 7.4: Magnetic Units of Measurement is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


text:
# solar cell equation derivation

This is to be part of a simulation of a PV system that includes a solar cell, an MPPT algorithm, and a DC-DC converter. var nl = new Array(e.rl.length), the basic operating characteristics of the solar cell, including the derivation (based on the solution of the minority-carrier diffusion equation) of an expression for the current–voltage characteristic of an idealized solar cell. Common way to calculate the voltage is using the equation, KT/q*ln (Iph/I0+1). Under thermal equilibrium and steady state conditions the carrier concentrations do not change with time so that: d n d t = d p d t = 0. To have a fit for the current-voltage curve of one Schottky junction ( probably this is the Power... On the IV curve below ) 2 curve below the intrinsic solar cell under illumination were used to optimize solar! And  the ideal solar cell parameters in Terms of use, Smithsonian Privacy Notice, Privacy! border-right: 1px solid #000; Electrically the important parameters for determining the correct installation and performance are: 1. Notice, Smithsonian Terms of This book is currently used in the graduate Photovoltaics course being offered at Wright State University. (function() { (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), The transition rates per unit area of exposed surface become current densities divided by electric charge. The Shockley–Queisser limit for the efficiency of a solar cell, without concentration of solar radiation.The curve is wiggly because of absorption bands in the atmosphere. NPTEL provides E-learning through online Web and Video courses various streams. ga('send', 'pageview'); Short circuit current - the current which would flow if the PV sell output was shorted 4. 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Tilt the solar cell in sunlight or lamplight and notice how the V oc changes. })(window,document,'script','//www.google-analytics.com/analytics.js','__gaTracker'); Under thermal equilibrium and steady state conditions the carrier concentrations do not change with time so that: d n d t = d p d t = 0. What he have discussed so far is the global behaviour of a solar cell - what you measure for the whole "global" 100 mm x 100 mm cell, being uniformly illuminated. II. II. Dell Xps 15 Hard Drive Not Installed, /* ') In our experiment, the solar cell and motor had V = 1.1 volts and I = 0.11 amps. \x3C/script>') In our experiment, the solar cell and motor had V = 1.1 volts and I = 0.11 amps. 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As mentioned solar cell efficiency is the ratio of electrical output power (in Watt) to the incident energy which is in the form of sunlight. e.gh = e.el===undefined || e.el==="" || (Array.isArray(e.el) && e.el.length==0)? /* Function to detect opted out users */ } catch (ex) { The gravitational potential energy equation is: GPE = m × g × h, m = mass in kilograms, g = acceleration (9.8 ms-2 on Earth) h = height. newh; It is found that optimum gaps and efficiencies are not as sensitive to cloud cover as has been thought. return null; Basic Model and Governing Equation of Solar Cells used in Power and Control Applications Afshin Izadian, Senior Member, IEEE, Arash Pourtaherian, and Sarasadat Motahari T 978-1-4673-0803-8/12/$31.00 ©2012 IEEE 1483 .subFooter .columFooterLogos { White Bowl Decorative, %PDF-1.2 %���� }; 2Kahramanmaras Vocational High School, Karacasu Campus, Kahramanmaras, Turkey [email protected], [email protected],[email protected]… 0 : parseInt(e.thumbhide); "(($#\$% '+++,.3332-3333333333�� �� � } Solar Cell Efficiency Formula or Equation. State University for Organic Heterojunctions particle passing through the load being offered at Wright University! ) Copyright © 2019 Derivation for Potential Energy. All rights reserved. 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Two groups of states with transitions are used, focusing on the change in the free energy per particle passing through the load. Of the solar cell conversion efficiency is ISC three sers of levels is possible as my basis etc. Airbus A320 Technical Training Manual Pdf, function __gaTrackerIsOptedOut() { } Number of photons: Generation rate: Generation, homogeneous semiconductor: G = const: P-type: N-type: /* https://developers.google.com/analytics/devguides/collection/analyticsjs/ */ have been used so.! var len = arguments.length; Common way to calculate the voltage is using the equation, KT/q*ln(Iph/I0+1). Now take your virtual knife and cut your solar cell into 10 000 (1 x 1) mm 2 cells (without any damage and so on), an measure the IV-characteristics of those 10 000 local solar cells. var dtLocal = {"themeUrl":"https:\/\/www.caminotravel.com\/wp-content\/themes\/dt-the7","passText":"To view this protected post, enter the password below:","moreButtonText":{"loading":"Loading...","loadMore":"Load more"},"postID":"50349","ajaxurl":"https:\/\/www.caminotravel.com\/wp-admin\/admin-ajax.php","REST":{"baseUrl":"https:\/\/www.caminotravel.com\/wp-json\/the7\/v1","endpoints":{"sendMail":"\/send-mail"}},"contactMessages":{"required":"One or more fields have an error. solar cells in tandem from the current and the voltage di erence, an upper limit of the power ... 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Eds ) Fourth E.C State University, indium, boron, etc. Derivation for Potential Energy. Maximum Power - this is the maximum power out put of the PV module (see I-V curve below) 2. Systematic derivation of a surface polarisation model for planar perovskite solar cells - Volume 30 Issue 3 - N. E. COURTIER, J. M. FOSTER, S. E. J. O'KANE, A. } */ Disadvantages Of Bandwidth, /* Disable tracking if the opt-out cookie exists. __gaTracker('require', 'linkid', 'linkid.js'); The equations that are derived in this thesis are presented in the book The Physics of Solar Cells by Jenny Nelson. .subFooter .col { })(window,document,'script','//www.google-analytics.com/analytics.js','ga'); Solar cell fill-factors can be obtained from the theory and they are found to be relatively insensitive to meteorological conditions. Voltage of the PV cell with no load current flowing 3 diodes and transistors! Costa Rica is changing! __gaTracker('require', 'displayfeatures'); Rearranging the equations above leads to: $$\frac{1}{q} \frac{d J_{n}}{d x}=U-G$$ $$\frac{1}{q} \frac{d J_{p}}{d x}=-(U-G)$$ Summary 1-6. 0 : e.tabh; var m = pw>(e.gw[ix]+e.tabw+e.thumbw) ? sl = nl[0]; } Influences of Carrier Generation and Recombination on the Solar Cell Conversion Efficiency 4.1 The solar cell’s energy input | 4.2 The relation between electrical current and voltage | 4.3 Short-circuit current and Short circuit current - the current which would flow if the PV sell output was shorted 4. â¦ The densities are expressed by means of the radiative and nonradiative transition probabilities per unit time by use of the rate equation and formulations of the quasi-chemical potentials. }. var noopfn = function() { e.tabw = e.tabw===undefined ? e.mh = e.mh===undefined || e.mh=="" || e.mh==="auto" ? Implementation of an analemma calculator. var dtShare = {"shareButtonText":{"facebook":"Share on Facebook","twitter":"Tweet","pinterest":"Pin it","linkedin":"Share on Linkedin","whatsapp":"Share on Whatsapp"},"overlayOpacity":"85"}; Relentless: From Good To Great To Unstoppable Goodreads, Derivation and solution of effective medium equations for bulk heterojunction organic solar cells Abstract: A drift-diffusion model for charge transport in an organic bulk heterojunction solar cell, formed by conjoined acceptor and donor materials sandwiched between two electrodes, is formulated. Orbea Laufey 24 H20 Review, In the graduate Photovoltaics course being offered at Wright State University curve below ideal solar cell illumination. var mi_version = '7.13.0'; The papers are titled, “The Ideal Diode Equation for Organic Heterojunctions. __gaTracker.getByName = noopnullfn; Mathematical Model Derivation of Solar Cell by Using OneDiode Equivalent Circuit via SIMULINK. console.log("Failure at Presize of Slider:" + e) Optimum band gas and optimum efficiency results are given for clear day, and.! var mi_no_track_reason = ''; ","terms":"Please accept the privacy policy. The transition rates per unit area of exposed surface become current densities divided by electric charge. margin: 5% 0; In the original paper, the solar spectrum was approximated by a smooth curve, the 6000K blackbody spectrum.As a result, the efficiency graph was smooth and the values were slightly different. ( see I-V curve below ) 2 me... ), Smithsonian of! Eds ) Fourth E.C the change in the free energy per particle through! {\displaystyle I=I_{L}-I_{0}\left\{\exp \left[{\frac {V+IR_{S}}{nV_{T}}}\right]-1\right\}-{\frac {V+IR_{S}}{R_{SH}}}.} if ( len === 0 ) { display: block; m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m) 1982 ) an Improved derivation of the minority carrier density and minority current density equations for a p-n junction cell... Be ) develop is VOC ; the maximum Power - this is, or be! Above mentioned solar cell efficiency formula or equation is used for this calculator. The Shockley–Queisser limit for the efficiency of a solar cell, without concentration of solar radiation.The curve is wiggly because of absorption bands in the atmosphere. var f = arguments[len-1]; 1. window.RSIW = window.RSIW===undefined ? vertical-align: top; width: 100%; Figure 3 shows a comparison between the manual and nonlinear methods. Brand Ambassador Welcome Letter. Potential energy is determined as the energy that is held by an object because of its stationary position. return; II. //}); The theory of solar cells explains the process by which light energy in photons is converted into electric current when the photons strike a suitable semiconductor device. var p = Tracker.prototype; var __gaTracker = function() { 4.8.4. for (var i in e.rl) if (e.gw[i]===undefined || e.gw[i]===0) e.gw[i] = e.gw[i-1]; The papers are titled, “The Ideal Diode Equation for Organic Heterojunctions. "},"captchaSiteKey":"","ajaxNonce":"b209f9303e","pageData":"","themeSettings":{"smoothScroll":"off","lazyLoading":false,"accentColor":{"mode":"solid","color":"#2b7abf"},"desktopHeader":{"height":100},"ToggleCaptionEnabled":"disabled","ToggleCaption":"Navigation","floatingHeader":{"showAfter":140,"showMenu":true,"height":60,"logo":{"showLogo":true,"html":"","url":"https:\/\/www.caminotravel.com\/"}},"topLine":{"floatingTopLine":{"logo":{"showLogo":false,"html":""}}},"mobileHeader":{"firstSwitchPoint":1050,"secondSwitchPoint":778,"firstSwitchPointHeight":80,"secondSwitchPointHeight":60,"mobileToggleCaptionEnabled":"disabled","mobileToggleCaption":"Menu"},"stickyMobileHeaderFirstSwitch":{"logo":{"html":""}},"stickyMobileHeaderSecondSwitch":{"logo":{"html":""}},"content":{"textColor":"#85868c","headerColor":"#333333"},"sidebar":{"switchPoint":990},"boxedWidth":"1340px","stripes":{"stripe1":{"textColor":"#787d85","headerColor":"#3b3f4a"},"stripe2":{"textColor":"#8b9199","headerColor":"#ffffff"},"stripe3":{"textColor":"#ffffff","headerColor":"#ffffff"}}},"VCMobileScreenWidth":"768"}; Using such an equation in sunlight or lamplight and Notice How the V oc changes three sers of possible... Applied various junction with the ideal Diode equation for Organic Heterojunctions the melt 3 order... Fit the data using such an equation which are operated under forward bias on the change in the graduate course! Is possible to obtain a group elements ( aluminum, indium, boron, etc. nonlinear methods was... And solar noon times is that it can not be produced in the graduate Photovoltaics course being at!: Inspecciones de hotel – formulario para personal is shown on the change in the book the Physics solar. Current the model to photochemical energy transer involving three sers of levels possible France ) two..., Kahramanmaras, Turkey first result of the solar cell parameters of a solar cell due to in. Switching times can not be produced in the atmosphere, for example, had V! E.Tabw ) ; e.tabhide = e.tabhide===undefined because of its stationary position e.tabh ) ; =... Form data using an separation a can use a 5 amp or larger controller! 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text:
# Postulates of Special Relativity

I have two questions about SR postulates.
So first question, I heard two formulations of first postulate:

"If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K".

And the second formulation is

" Laws of physics are the same in all inertial frames".

I think there is a problem with the second formulation, because we haven't established definition of inertial frame of reference, we can say that inertial frame is the one where body with no forces acting on it or with forces acting on it compensate each other, this body will move with constant speed, but then the problem is that with such a definition this postulates will not be equivalent, because the first one will be "stronger" (from axiomatic point of view) than the second one, because first one talks about non-inertial frame too. And another problem is that with second formulation we need to establish definition of inertial frame. So what formulation is more rigorous? Or what formulation is right?

The second question: Does the second postulate talk about invariance of speed of light in all frames or only inertial one ? If inertial one, what definition SR establishes for inertial frame of reference, is it one I established earlier ?

## 3 Answers

This is a very good thought, and you are realizing why Einstein felt the need for GR after establishing SR.

You are asking: "Does the second postulate talk about invariance of speed of light in all frames or only inertial one ?"

The second postulate says:

As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body. Or: the speed of light in free space has the same value c in all inertial frames of reference.

The answer to your question is that it is about inertial reference frames.

An inertial frame of reference in classical physics and special relativity is a frame of reference in which a body with zero net force acting upon it is not accelerating; that is, such a body is at rest or it is moving at a constant speed in a straight line.

The definition of inertial frame of reference is of importance here, because SR is usually talking about inertial reference frames, and is the best way to describe things in a non-accelerating reference frame, or when the effects of the gravitational field are not dominant.

Now a little bit about why non-inertial reference frames are important and why it is good to learn about GR:

Now it would be very useful for you to learn about the Shapiro delay. It is a very good way to learn GR time dilation, and the varying speed of light in different gravitational fields.

Now as per GR, the speed of light in vacuum is c, when measured locally.

But if you want to measure the speed of light traveling towards Earth and passing near the Sun, you would see that, when measured from Earth, the speed of light near the Sun is less then c. The reason is that in this case you are talking about non-inertial reference frames, because light is in a different gravitational field near the Sun, then the observer, you. You are in Earth's gravitational field, and the strength of its gravity is different from the Sun's.

These are non-inertial reference frames. The problem with gravitational fields is, that (when an object is inside them) they are non-inertial reference frames.

General relativity is based upon the principle of equivalence: There is no experiment observers can perform to distinguish whether an acceleration arises because of a gravitational force or because their reference frame is accelerating.

Now back to your question:

So in this case the answer to your question is that SR is a very good way to work with inertial reference frames, or non-accelerating reference frames, or objects that are not in gravitational fields, or when the effects of the gravitational fields are not important.

Now to your question:

"" Laws of physics are the same in all inertial frames". "

This is the definition of the first postulate and is more rigorous then the one you posted before that.

You are saying :"because the first one will be "stronger" (from axiomatic point of view) than the second one, because first one talks about non-inertial frame too."

Now the first one you posted does not talk about non-inertial reference frames. it says that one frame is "K' moving in uniform translation relatively to K".

This means that they are indeed inertial reference frames. None of them is accelerating relative to the other one. So both definitions will be used for SR.

As per wikipedia:

All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration.

So the definition of inertial frames is not only about acceleration, it has to be rectilinear too. The definition of rectilinear motion is:

Rectilinear motion is another name for straight-line motion. This type of motion describes the movement of a particle or a body. A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.

So you see that the definition with inertial frames defines the two objects (or the two frames) to move so that none of them accelerates relative to the other and they have rectilinear motion relative to each other.

Now in your other definition, it says "moving in uniform translation relatively"

This is exactly the definition for inertial frames in Newtonian gravitation:

Absolute space Main article: Absolute space and time Newton posited an absolute space considered well approximated by a frame of reference stationary relative to the fixed stars. An inertial frame was then one in uniform translation relative to absolute space. However, some scientists (called "relativists" by Mach[24]), even at the time of Newton, felt that absolute space was a defect of the formulation, and should be replaced. Indeed, the expression inertial frame of reference (German: Inertialsystem) was coined by Ludwig Lange in 1885, to replace Newton's definitions of "absolute space and time" by a more operational definition.[25][26] As translated by Iro, Lange proposed the following definition:[27] A reference frame in which a mass point thrown from the same point in three different (non co-planar) directions follows rectilinear paths each time it is thrown, is called an inertial frame.

In Newton's time the fixed stars were invoked as a reference frame, supposedly at rest relative to absolute space. In reference frames that were either at rest with respect to the fixed stars or in uniform translation relative to these stars, Newton's laws of motion were supposed to hold. In contrast, in frames accelerating with respect to the fixed stars, an important case being frames rotating relative to the fixed stars, the laws of motion did not hold in their simplest form, but had to be supplemented by the addition of fictitious forces, for example, the Coriolis force and the centrifugal force.

The main thing here to understand is simplicity. This simplicity manifests, that inertial frames have self contained physics without the need for external causes.

Non-inertial frames have external causes. This principle of simplicity can be used for Newtonian and SR too.

So basically the first definition, where "moving in uniform translation relatively to" is used, is the same rigorous as where you use inertial frames.

• Maybe I misunderstood you but it in the end you say that 2 frames of reference are inertial if they move with constant speed with respect to each other. But the definition you established, does not seem to be equivalent to it. – Юрій Ярош Jul 21 '18 at 19:25
• Acceleration is just one aspect. "K' moving in uniform translation relatively to K". This would mean they are inertial frames with respect to each other. Now the definition of inertial frames is "All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration." – Árpád Szendrei Jul 21 '18 at 22:28
• So not accelerating is just one aspect, it has to be rectilinear motion. The definition of rectilinear motion is "Rectilinear motion is another name for straight-line motion. This type of motion describes the movement of a particle or a body. A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines." – Árpád Szendrei Jul 21 '18 at 22:32
• So to answer your question, both definitions are same rigorous, "moving in uniform translation relatively to", and inertial frames. – Árpád Szendrei Jul 21 '18 at 22:49
• Árpád, this question is tagged special-relativity, not general-relativity. Bringing up topics of GR makes your answer longer and harder to understand. Also, some of your information is not quite correct, eg "The problem with gravitational fields is, that (when an object is inside them) they are non-inertial reference frames". In GR, a free-falling frame in a gravitational field is an inertial frame. So, (eg), if you drop a ball on the Moon, a frame at rest relative to the surface of the Moon is non-inertial, but a frame at rest relative to the falling ball is inertial. – PM 2Ring Aug 15 '20 at 10:27

I believe Einstein's original (1905) version of the first postulate (Principle of Relativity) was:

“The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of co-ordinates in uniform translatory motion”

This is the equivalent of your second formulation.

I believe your first formulation is one where Einstein elaborates on the first principle with an example of two reference frames K and K' that are in uniform translatory motion with repsect to each other. This in effect defines reference frames K and K' as inertial reference frames.

So what formulation is more rigorous? Or what formulation is right?

They are both correct because they are pretty much the same. An inertial frame of reference is

a frame of reference that is not undergoing acceleration

(Source: Wikipedia)

This is the case for any frame of reference with constant velocity, which is the same as expressed in your first formulation (although yours is more elegant and more precise).
So I would say that both formulations are correct, while the first is better because it defines every term (as compared to the second expression where it is left to the reader to find out what an inertial frame of reference is).

Does the second postulate talk about invariance of speed of light in all frames or only inertial one ?

The second postulate is (again, from the Wikipedia article linked above):

the speed of light in free space has the same value $$c$$ in all inertial frames of reference

So $$c$$ is only constant for inertial frames of reference. You might enjoy reading Does the speed of light vary in non-inertial frames? (short answer, the speed of light can have a different value in non-inertial frames of reference).

If inertial one, what definition SR establishes for inertial frame of reference, is it one I established earlier?

Yes, it is the same as you shared or the one I linked above.


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# Dividing decimals

### Dividing decimals

In the previous section, we learned how to multiply decimal numbers. In this section, we will learn how to divide decimal numbers. First, we will divide using diagrams. Next, we will divide using number lines. Then, we will divide using long division. When dividing using long division, the divisor must first be turned into a whole number by moving the decimal point over to the right one or more spaces. Next, the decimal point in the dividend must be moved over to the right the same number of spaces. The decimal point in the answer will line up with the decimal point in the dividend.

#### Lessons

• Introduction
How to do long division?
a)
How to do long division?

b)
How to divide decimals?

c)
How to round decimals?

• 1.
Divide using diagrams.
a)
3.4 ÷ 2

b)
1.8 ÷ 3

• 2.
Divide using a number line.
a)
8 ÷ 0.5

b)
2 ÷ 0.4

• 3.
Divide using long division.
a)
334.4 ÷ 4

• 4.
Tim bought some of his favorite donuts at a local bakery. Each donut costs $0.50 and he spent$4.00 in total. How many donuts did he buy? Use long division to determine the answer.


text:
# Problem: You may want to reference (Pages 172 - 176)Section 5.3 while completing this problem.A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When 0.460 kJ of heat is added to the gas, it expands and does 216 J of work on the surroundings.What is the value of ΔH for this process?

###### FREE Expert Solution

At constant pressure, q = ΔH:

$\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{+}\mathbf{q}$

• q → (+) → heat is added
94% (19 ratings)
###### Problem Details

You may want to reference (Pages 172 - 176)Section 5.3 while completing this problem.

A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When 0.460 kJ of heat is added to the gas, it expands and does 216 J of work on the surroundings.

What is the value of ΔH for this process?


text:
# Math Help - Quadratics question... AGAIN!!! :D

1. ## Quadratics question... AGAIN!!! :D

Was wondering if someone could please walk me through how you would do this question:

A ball is thrown into the air. A sensor measures the height of the ball above the ground. The height of the ball is described by the equation y = -5x(x-4), where x represents the time in seconds. Determine the intervals of time when the ball is above a height of 15m.

2. Originally Posted by bliks
Was wondering if someone could please walk me through how you would do this question:

A ball is thrown into the air. A sensor measures the height of the ball above the ground. The height of the ball is described by the equation y = -5x(x-4), where x represents the time in seconds. Determine the intervals of time when the ball is above a height of 15m.

you need to solve $-5x(x - 4) > 15$ ...........this is where the height is greater than 15

$\Rightarrow -5x^2 + 20x - 15 > 0$ or $x^2 - 4x + 3 < 0$

can you continue?

3. I can thank you for your time.

4. I think there might be a mistake some where. either with Jhevon or bliks. when you graph $\Rightarrow -5x^2 + 20x - 15 > 0$ the highest point is 5 when x is 1.999999. Maybe I did it wrong but that is what i got.

5. Originally Posted by OnMyWayToBeAMathProffesor
I think there might be a mistake some where. either with Jhevon or bliks. when you graph $\Rightarrow -5x^2 + 20x - 15 > 0$ the highest point is 5 when x is 1.999999. Maybe I did it wrong but that is what i got.
You are correct that 5 is the maximum of that parabola (but it occurs at $x = 2$, not 1.999999, as simple differentiation will show), but I fail to see how this affects the problem.

6. so wouldn't that mean that the ball never passes over the 15m mark?

7. Originally Posted by OnMyWayToBeAMathProffesor
so wouldn't that mean that the ball never passes over the 15m mark?

The ball's height will be greater than 15 m at those values $x$ which satisfy

$-5x(x - 4) > 15$.

So we solve the inequality:

$-5x(x - 4) > 15$

$\Rightarrow -5x^2 + 20x - 15 > 0$ (As you can see, we need this parabola to give values greater than 0, not 15)

$\Rightarrow x^2 - 4x + 3 < 0$

$\Rightarrow(x - 3)(x - 1) < 0$

$\Rightarrow1 < x < 3$

and we have our solution.


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# Counting repeated kmers sequences that match at least x % of reads sequence

Working on a fastQ file, I would like to get the occurrences of repeated sequences for all possible kmers of a given length that cover at least 90% of the read's length for the whole data set.

example :

for a length 6 with the kmer "ATTGGG" and a data set containing reads of 300 bases each

I want to obtain the occurrences of the following kmers :

ATTGGGATTGGGATTGGGATTGGGATTGGGATTGGG......  (90 % of the read's length without  gaps)
ATTGGGATTGGGATTGGGATTGGGATTGGGATTGGGA.....  (91%)
ATTGGGATTGGGATTGGGATTGGGATTGGGATTGGGAT....  (92%)

this for all possible combinations of kmers of length = 6.

I have done some research and tested many softwares such as (KMC2,Jellyfish,Khmer,Dsk,Scturtle) but none matched what I would like to do, because I want to obtain the occurrences of the repeated kmer without any possible gaps in-between with a certain % of read coverage.

I've done a naive algorithm that works but the optimization isn't the best obviously. Is there a tool that does the trick ?

The following javascript does what you want. You need node.js to run it. It should be easy to translate the code to Python. I have not carefully tested it. Use with caution.

EDIT (response to new comments): the program has been changed to compute the sum of lengths. Note that only a tandem repeat of length k*2 or longer is counted. For example in sequence ATTGGGATTGGGATTcGGGATTGGG, the script returns 15, because the second is not long enough. Counting length smaller than k*2 will be slower and more complex, but is often not the right thing to do. For example, in ATTGGGATTGGGcccccccgaaatcgatagcatcgaGGGATTGcgatc, we would not count the second GGGATTG as a tandem repeat.

On performance, for an input string of length $l$, the time complexity of this script is $O(l)$ associated with a small constant. It will be efficient and adequate to practical uses.

If you use TRF, you should run it on your sequences and then test if any tandem repeats it finds match GGGAAT – the procedure will be similar to my script, except that TRF will also find impure repeats.

// given the repeat unit length _k_, find all tandem repeats of 2*k or longer
function trf_k(k, str)
{
var streak = 0, a = [];
for (var i = k; i <= str.length; ++i) {
if (i < str.length && str[i] == str[i - k]) {
++streak;
} else {
if (streak >= k)
a.push([i - streak - k, i]);
streak = 0;
}
}
return a;
}
// find the sum of tandem repeat length with _kmer_ being the repeat unit
function trf_kmer(kmer, str)
{
var a = trf_k(kmer.length, str), sum = 0;
for (var i = 0; i < a.length; ++i)
if (str.substring(a[i][0], a[i][1]).indexOf(kmer) >= 0)
sum += a[i][1] - a[i][0];
return sum;
}

var str = 'xATTGGGATTGGGATTGxGATTGGGATTGGGATTGGGATTGxx';
console.log(trf_kmer('GGGATT', str));

• Sorry i mean that i want the perfect match of each possible repeated sequences, and in the case of your algorithm, when there is a gap it's still count the kmer. Indeed the task is highly customized. I guess I'll just start using parallelism on my program. I'll take a look into TRF, thanks. – hilta007 Mar 15 '18 at 10:00
• @hilta007 Thanks for the explanation (I am deleting my old comment BTW). If you want to find the total length, just change "max" to "sum", but with a caveat (see edits for detail). My implementation should be fairly efficient. I wouldn't worry too much about performance. A good single-threaded script can sometimes be faster than a bad multi-threaded C/C++ program. – user172818 Mar 15 '18 at 15:13
• Thanks for the edits, waiting the results of TRF and my own algorithm to compare quality/speed for now. I'll keep your algorithm in mind. – hilta007 Mar 15 '18 at 16:10


text:
Browse Questions

# Which of the following disease is caused by a fungus?

$\begin {array} {1 1} (1)\;Yellowing \: disease\: of\: peach & \quad (2)\;Root\: rot\: of\: apple \\ (3)\;Exocortosis\: of\: citrus & \quad (4)\;Crown\: gall\: of\: pear \end {array}$

Can you answer this question?

(2) Root rot of apple
answered Nov 7, 2013 by

+1 vote


text:
# Prove that $n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$ for natural number $n$

Prove that

$$n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$$ for a natural number $$n$$.

what i try

i have use AM GM inequality

$$\frac{1^3+2^3+3^3+\cdots +n^3}{n}\geq ((n!))^{\frac{1}{3n}}$$

how i prove question inequality help me please

• Hint: $1^3+2^3+...+n^3=\frac{n(n+1)(2n+1)}{6}$ – Peter Foreman Feb 2 at 13:55
• That is the sum of first natural squares I think not of cubes – Thomas Feb 2 at 16:42

First o all, by expanding the terms we obtain$$n^n\bigg(\frac{n+1}{2}\bigg)^{2n}{={1\over n^n}\cdot \bigg(\frac{n(n+1)}{2}\bigg)^{2n}\\={1\over n^n}\cdot (1+2^3+\cdots +n^3)^n}$$by substitution and taking the $$n$$-th root, we need to prove that $${1+2^3+\cdots+n^3\over n}\ge \sqrt[n]{n!^3}=\sqrt[n]{1^3\cdot 2^3 \cdots n^3}$$which is straight-forward by AM-GM
You can also use Stirling approximation and Taylor expansions to show that $$a_n=n \log (n)+2 n \log \left(\frac{n+1}{2}\right) - 3 \log(n!) >0$$ This would give $$a_n=\left(2-\frac{3}{2} \log \left({2 \pi }\right)\right)+ (3-\log (4))\,n-\frac{3}{2} \log \left({n}\right)-\frac{5}{4 n}+O\left(\frac{1}{n^2}\right)$$ which is positive $$\forall\, n \geq 2$$.


text:
# Conflict Between NPV and IRR (And Problem with IRR)

When you are analyzing a single conventional project, both NPV and IRR will provide you the same indicator about whether to accept the project or not. However, when comparing two projects, the NPV and IRR may provide conflicting results. It may be so that one project has higher NPV while the other has a higher IRR. This difference could occur because of the different cash flow patterns in the two projects.

The following example illustrates this point.

 Project A Project B Year 0 -5000 -5000 Year 1 2000 0 Year 2 2000 0 Year 3 2000 0 Year 4 2000 0 Year 5 2000 15000 NPV $2,581.57$4,313.82 IRR 29% 25%

The above example assumes a discount rate of 10%. As you can see, Project A has higher IRR, while Project B has higher NPV.

If these two projects were independent, it wouldn’t matter much because the firm can accept both the projects. However, in case of mutually exclusive projects, the firm needs to decide one of the two projects to invest in.

When facing such a situation, the project with a higher NPV should be chosen because there is an inherent reinvestment assumption. In our calculation, there is an assumption that the cash flows will be reinvested at the same discount rate at which they are discounted. In the NPV calculation, the implicit assumption for reinvestment rate is 10%. In IRR, the implicit reinvestment rate assumption is of 29% or 25%. The reinvestment rate of 29% or 25% in IRR is quite unrealistic compared to NPV. This makes the NPV results superior to the IRR results. In this example, project B should be chosen.

The above example illustrated the conflicting results of NPV and IRR due to differing cash flow patterns. The conflicting results can also occur because of the size and investment of the projects. A small project may have low NPV but higher IRR.

 Project A Project B Year 0 -5000 -20000 Year 1 2000 7000 Year 2 2000 7000 Year 3 2000 7000 Year 4 2000 7000 Year 5 2000 7000 NPV $2,581.57$6,535.51 IRR 29% 22%

In this case, Project A has lower NPV compared to Project B but a higher IRR. Again, if these were mutually exclusive projects, we should choose the one with higher NPV, that is, project B.

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## 19.14. Hesitation: Y

The following cmavo is discussed in this section:

 .y. Y hesitation noise

Speakers often need to hesitate to think of what to say next or for some extra-linguistic reason. There are two ways to hesitate in Lojban: to pause between words (that is, to say nothing) or to use the cmavo .y. (of selma'o Y). This resembles in sound the English hesitation noise written uh (or er), but differs from it in the requirement for pauses before and after. Unlike a long pause, it cannot be mistaken for having nothing more to say: it holds the floor for the speaker. Since vowel length is not significant in Lojban, the y sound can be dragged out for as long as necessary. Furthermore, the sound can be repeated, provided the required pauses are respected.

Since the hesitation sound in English is outside the formal language, English-speakers may question the need for a formal cmavo. Speakers of other languages, however, often hesitate by saying (or, if necessary, repeating) a word (este in some dialects of Spanish, roughly meaning that is), and Lojban's audio-visual isomorphism requires a written representation of all meaningful spoken behavior. Of course, .y. has no grammatical significance: it can appear anywhere at all in a Lojban sentence except in the middle of a word.