Elastic fibers (or yellow fibers) are an essential component of the extracellular matrix composed of bundles of proteins (elastin) which are produced by a number of different cell types including fibroblasts, endothelial, smooth muscle, and airway epithelial cells. These fibers are able to stretch many times their length, and snap back to their original length when relaxed without loss of energy. Elastic fibers include elastin, elaunin and oxytalan.
Elastic tissue is classified as "connective tissue proper".Elastic fibers are formed via elastogenesis, a highly complex process involving several key proteins including fibulin-4, fibulin-5, latent transforming growth factor β binding protein 4, and microfibril associated protein 4 In this process tropoelastin, the soluble monomeric precursor to elastic fibers is produced by elastogenic cells and chaperoned to the cell surface. Following excretion from the cell, tropoelastin self associates into ~200 nm particles by coacervation, an entropically driven process involving interactions between tropoelastin's hydrophobic domains, which is mediated by glycosaminoglycans, heparan, and other molecules. These particles then fuse to give rise to 1-2 micron spherules which continue to grow as they move down from the cells surface before being deposited onto fibrillin microfibrillar scaffolds.Following deposition onto microfibrils tropoelastin is insolubilized via extensive crosslinking by members of the lysyl oxidase and lysyl oxidase like family of copper-dependent amine oxidases into amorphous elastin, a highly resilient, insoluble polymer that is metabolically stable over a human lifespan. These two families of enzymes react with the many lysine residues present in tropoelastin to form reactive aldehydes and allysine via oxidative deamination.These reactive aldehydes and allysines can react with other lysine and allysine residues to form desmosine, isodesmosine, and a number of other polyfunctional crosslinks that join surrounding molecules of tropoelastin into an extensively crosslinked elastin matrix. This process creates a diverse array of intramolecular and intermolecular crosslinks These unique crosslinks are responsible for elastin's durability and persistence. Maintenance of crosslinked elastin is carried out by a number of proteins including lysyl oxidase-like 1 protein.Mature elastic fibers consist of an amorphous elastin core surrounded by a glycosaminoglycans, heparan sulphate, and number of other proteins such as microfibrillar-associated glycoproteins, fibrillin, fibullin, and the elastin receptor.
So I understand that I have to integrate the negative of the force function to get the change in PE. I get -(20x^2 - 2x^3) and when I evaluate it from 0 to 2, I get -64N. But, of course, the change is positive. What am I missing?
Thank you.
Hello All :
i am reviewing charlse kettle book introduction to solid state physics and came across an equation in the 3ed chapter which is a bit unclear
hope that physics forums members can clear it more if that possible
the equation 1st mentioned in the 3ed chapter page number 84
it is about...
Hi,
Here is the problem
What is required to answer this question is two assumptions. Firstly, the component of the momentum normal to the centre line is the same before and after. Therefore, secondly, A must recoil entirely in the horizontal plane. This is the only way to answer this question...
I found that 1/2m1v1i^2+1/2m2v2i^2=1/2m1v1f^2+1/2m2v2f^2
=>0.5*200*55^2+0.5*46*0^2=0.5*40^2*200+0.5*46*0*vf^2=>vf=78.713 m/s.
The true answer is 65.2 m/s and is solved using m1v1i+mvv2i=m1v1f+m2v2f. Are these equations not interchangeable? Why can I not use the equation I used?
Equation 1 is equating the kinetic energies of the objects before and after the elastic collision. Equation 2 is equating the momentums of the objects after the elastic collision. They can be used interchangeably as long as the collision is elastic.
Am I right in my conclusion?
Teacher described the Thomson scattering effect through the lens of the electric field changing as a moving particle is accelerated. The changing electric field of the electron accelerating carries with it an amount of energy, and this energy radiates out from the acceleration event. (there were...
For this problem,
The solution is,
However, is the reason why they don't include electrical potential energy because the time interval for which we are applying conservation of energy over is very small so the change in electric potential energy is negligible?
Also, when they said, "electrons...
a) Elastic potential energy stored in the compressed spring is written by, where k =400N/m, compressed spring distance x = 0.5m
$$ U_g = \frac {1}{2}kx^2$$
$$ U_g = 50J$$
b) When block C is compressed, it has stored spring PE and when it is released, the block accelerates to the right, where it...
Hello,
so we have two potitions right, if we take ##\theta = 90## as the first position (i.e. both rods are flat) and then the second position at ##\theta = 0##.
I totally understand the exercise, not difficult. The only issue I am having is the torsional spring... it says that it is uncoiled...
Using principle of conservation of momentum:
m×u=m×v1 + M×v2
Where m=mass of moving particle in the beginning
u=Initial velocity of particle m
v1= final velocity of particle m
v2=velocity of object M
m×u-(mv1)=Mv2
(mu-mv1)÷M=v2
My answer is this (mu-mv1)÷M
However, it is nowhere close to...
I'm a little confused as to what the answer could be. This was one of my homework questions that I got wrong as I chose 0.5 v as the answer. Would someone be able to tell me what the correct answer would be?
In electron microscopy of thin solid specimens elastic scattering is treated as the main process responsible for formation of (phase contrast) images and diffraction patterns.
However, if an electron changes direction it should lose energy by producing a breaking radiation photon.
How can it be...
Hi,
Looking for the Elastic Constants for any rubber-like material such as Natural Rubber. It can be inorganic or organic. The constants I am looking for take the form of a fourth-rank tensor. I only need the first order elasticities, not the zeroth or higher (not Cij or Cijklmn.. just Cijkl)...
Suppose I have two spheres in 3 dimensions of equal mass. In cartesian coordinates, sphere A is traveling with velocity uAi, and sphere B travels with vBi. They will collide elastically.
I want to find the final velocities after the collision, ie uAf and vBf.
Am I correct in saying that...
From conservation of momentum:
m1u1 + m2u2 = m1v1 + m2
u1 - u2 = v1 + v2 (u2 is negative because the object moves to the left)
From conservation of KE, I got answer (C)
So there are two correct answers, (B) and (C)?
Thanks
Suppose a wooden cube with a side of 10 cm is connected to the bottom of a water container by a spring. If the density of wood is 0.6g/cm^3 and the density of water is 1g/cm^3 , what is the elastic force of the spring? Is it 4N or 6N ?
I think the answer is 4N … But the book says it is 6N ...
Is it currently possible to create a Graphene or Graphene Oxide-based polyester film, but with significantly lower helium permeability than currenlty avalible materials?
What would be the best way to create such a material and what type of processes and equipment would be involved in the...
I think the answer is that the elastic potential energy will be a 1/16th of the original value. This is my reasoning:
1) If the diameter doubles, the cross sectional area is 4 times the original value. (from A= πr2).
2) F= stress/area. Force (load is the same). If cross sectional area...
I started by writing the equation v1i + v1f = 2v and then drawing a triangle with v1i, v1f, and 2v as the three sides. Then I used the Law of Cosines to solve for cos theta but this did not lead to a solution. Could I have a hint on how to begin? Thank you!
Point B is elastic limit and point C is yield point.
From this link: https://en.m.wikipedia.org/wiki/Yield_(engineering)#Definition
The definition given is:
Both seems to refer to same definition, it is the point where the elastic deformation ends and plastic deformation begins. But from...
Sum of forces in the y-direction = 0 and downwards is +ve
P + Fab,y = 0
P + Fab (4/5) = 0
Fab = -1.25P
ẟ = FL/AE -> ẟab = FabLab/AabE
ẟab = (-1.25P*.75)/(pi*(.01)^2*(200*10^3)) = -0.0149P
After this step, I am uncertain of how I can relate the vertical elongation with AB's elongation to find...
I was able to solve for the velocity of MB and got my answer as 4.47m/s.
The main issue right now for me is how to get the angles. I'm really confused on what most people have been posting as we didn't get a groundwork on this topic and so most of the basics I had them self taught.
So far I...
Hi, I have some soft body equations that require first order elasticity constants. Just trying to figure out the proper indexing.
From Finite Elements of Nonlinear Continua by J.T. Oden, the elastic constants I am trying to obtain are the first order, circled below:
My particular constitutive...
Standard formula for final velocities ##v_1##, ##v_2## in elastic collision with masses ##m_1##, ##m_2## and initial velocities ##u_1##, ##u_2## is given by $$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2$$$$v_2 = \frac{2m_1}{m_1+m_2}u_1+\frac{m_2-m_1}{m_1+m_2}u_2$$.
By rearranging...
1 = elephant
2 = fly
So I am trying to find v'2 which is the final velocity of the fly. I have v1 the initial velocity of the elephant 2.1m/s. So I plug it into the equation and have v'2=(2m1/(m1+m2))*2.1m/s. We are not given the masses so I just know m1>m2 but I don't understand how that will...
Could I please ask for help with the following question?
Four uniform rods of equal length l and weight w are freely jointed to form a framework ABCD. The joints A and C are connected by a light elastic string of natural length a. The framework is freely suspended from A and takes up the shape...
I attempted to do mvf-mvi to find the impulse, but had trouble figuring out what to use as v (where does the angle of 3degrees come in?), and thought that there had to be more to the problem considering the other details I was given. I then attempted to maybe calculate the kinetic energy lost...
Hello,
I am wondering why in all X-ray diffraction experiments used to probe or know the crystal structure of the solid they assume that the scattering process is elastic, e.g, if an X-ray with wave vector ##k\vec{n}## is incident on a sample, it will diffract with a wave vector ##k\vec{n}'## of...
My approach so far is to use F = ma.
The forces acting on the block in the horizonital direction are friction and the force of the spring. Choosing the direction towards the spring as the positive axis.
Therefore: F = ma
-Fr - kx = ma
Solving for a = (-Fr - kx)/m
If I plug in values I end up...
The speed of the block after the nth collision is
$$ V_n=(2e)^n*v_0 $$
By conservation of energy the block travels a distance $$V_n^2/(2ug)$$ on the nth bounce. So the total distance is
$$ d=1/(2ug)∗(v_0^2+(2ev_0)^2...) $$
$$ d=1/(2ug)∗(v_0^2/(1−4e^2)) $$
$$ d=1/(2ug)∗(v_0^2∗M^2/(M^2−4m^2))...
Hi everyone,
I'm trying to understand the rationale behind the boundary condition for the problem "Finite bending of an incompressible elastic block". (See here from page 180).Here we have as Cauchy Stress tensor (see eq. (5.82)):
##T = - \pi I + \mu (\frac{l_0^2}{4 \bar{\theta}^2 r^2} e_r...
I am studying the finite bending of a rubber-like block, assuming Neo-Hookean response. In the following, ##l_0##,##h##, ##\bar{\theta}## are parameters, while the variables are ##r## and ##\theta##.
The Cauchy stress tensor is
##T= - \pi I + \mu(\frac{l_0^2}{4 \bar{\theta}^2 r^2} e_r \otimes...
I'm studying elasticity from classical Gurtin's book, and my professor gave us the following example, during lecture. Unfortunately, this is not present in our references, so I'm posting it here the beginning of the solution, and I will highlight at the end my questions. First I need to state...
Consider the system of the mass and uniform disc.
Since no external forces act on the system, the angular momentum will be conserved. For elastic collision, the kinetic energy of the system stays constant.Measuring angular momentum from the hinge:
##\vec L_i = Rmv_0 \space\hat i + I \omega_0...
https://www.plasmaphysics.org.uk/collision2d.htm
This is the only one I found, but when I plug in the numbers of his example I get a wrong result. Do you know any others who solved it i.e. considering the angle of impact? Angle of impact I name the angle that is shaped between the initial dx...
The way I learned to solve this was to switch to a frame of reference where one object is stationary.
given: m1 =0.6kg v1 = 5.0m/s [W], m2 = 0.8kg v2 = 2.0 m/s [E]
Setting v2 to rest by adding 2.0 m/s W to each object
New velocities are v1 = 7.0 m/s [E] and v2 = 0.0m/s
Then using the...
https://en.wikipedia.org/wiki/Elastic_collision
μα+mβ=μx+my,
μα^2+mβ^2=μx^2+my^2
I want x in relation of all variables except y, therefore I need to replace-eliminate y:
μα+mβ=μx+my =>y=(μα+mβ-μx)/m
μα^2+mβ^2=μx^2+my^2=>y=((μα^2+mβ^2-μx^2)/m)^0.5
and it is eliminated if I equate these two parts...
I am trying to program a game in which I have a bouncy rubber ball which upon collision with the ground, will have a squeeze factor (like a spring) and will bounce back higher due to spring physics.
I have currently been able to make a rigid bouncing ball since that simply follows the rules of...
1. Hello, so the difficulty I am having with this problem is that is seems relatively straightforward. I have tried to solving it by assuming that this is a collision in which momentum is conserved. Therefore, I found the total momentum before the collision and used this to resolve it must be...
In every book I checked, the energy (per unit mass) of elastic deformation is derived as follows:
## \int \sigma_1 d \epsilon_1 = \frac{\sigma_1 \epsilon_1}{2} ##
and then, authors (e.g. Timoshenko & Goodier) sum up such terms and substitute ##\epsilon ## from generalised Hooke's law i.e.
##...
So I've managed to confuse myself on this problem :)
Since the problem says we can assume ##m_p << m_b##, I'm assuming that the velocity of the bowling ball will be unchanged, such that ##\vec v_{b,i} = \vec v_{b,f} = -v_{b,0} \hat i##
I started out using the energy-momentum principle, ##(\vec...
Problem 2 – Composites
A composite component (such as shown in Fig.2) is required for an aerospace application. The specification for the component stipulates that it must have an Elastic Modulus in the fibre direction of at least 320 GN/m2, and the transverse direction modulus must not be less...
Since in an elastic collision, both momentum and energy is conserved,
P(initial)=P(final)
m1(3v)=m1v+m2v
m2/m1=2
Which was the given answer but if we use conservation of energy,
K.E(initial)=K.E(final)
1/2*m1*(3v)^2=1/2*m2*v^2+1/2*m1*v^2
m2/m1=8
Why do we get two different answers and why...
a)plastic deformation because of permanent deformation
b) the other parts that have been destroyed have stored the energy and this saved the passenger compartment.
C) the alloy crash barrier is stronger than the car body and and saves more of the energy by deforming shape.
I'm not sure about my...
So to cut to the chase, I missed my class' lesson on momentum - have tried to catch up, quite successfully but am baffled about this question. I know the conservation of momentum etc. but after trying for ages it's just not happening this question so any help would be much appreciated,
Oscar.
1.p=mv
Before the collision:
p hydrogen = 1.7x10^-27 * 500 =8.5*10^-25 kg ms^-1
p carbon = 2.0x10^-26 * 0 = 0 kg ms^-1
p total before = 8.5*10^-25 kg ms^-1
The sum of momentum prior to the collision is equal to the total momentum after a collision, momentum is constant, therefore;
p before = p...
My guess is that the deformation immediately before would be the sum of elastic and plastic deformation, and the deformation after would be just the plastic deformation, and it therefore would decrease. Is this correct?
After simplifying the equations, I got:
m1(v1-v1') = m2v2' (momentum) and
m1(v1-v1')(v1+v1') = m2v2'^2 (kinetic energy)
From there, I'm not sure what to do. I referred to a textbook and it said to divide the energy equation by the momentum equation (the simplified versions) and then do a...