why the static coefficient is more than kinetic coefficient?
445341
The static coeffiecent is larger simply because it is harder to move an object on a surface from rest than to keep moving the same object on a surface. That is all the coefficent is is an observed value for a certain object on a specific surface.
1132602
something that is static is not moving. For examle I am static when I am in my chair. Um.. Static Friction is the force of friction applied to an object when it has no velocity relative to whatever it is sitting on.\n\nVelocity dx/dt. Um... its speed but it includes direction! \n\nTotal Pressure, depends in what terms you are look at. Pressure is just Force/Area. Then you have pressure from a gas and then a liquid and then internal pressure.... all sorts of fun stuff!
816105
Generally, some of their collective kinetic energy is converted into heat, so the KE after the collision is less than the initial KE.  Momentum is conserved.
863216
It has to do more with the material of the ground rather than with the surface. It all depends on it's elastic coeficient. The least elastic, the more it will bounce because of the transmission of the kinetic energy.
170255
It is much easier to just give you a link instead of explaining it:\n\nhttp://www.shef.ac.uk/physics/teaching/phy107/solvegray.html
660156
http://www.physlink.com/education/askexperts/ae25.cfm
269520
Yeah, what do you mean by "statics"? Be more specific.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.
771396
Because gravity is the same - so it has the same effect on both objects.  The larger object will strike the ground with a greater force, but both objects will accelerate at the same rate.
44667
You have a contradiction in your statement.  Static friction implies no movement but velocity is movement.  \n\nIf you mean the velocity of an object in dynamic equilibrium has a result of a frictional force (kinetic though) like sliding a block across the floor then you have to know the amount of work put into the system.  Then use work-kinetic energy theorem.
48671
Static Friction\nStatic friction is some thing that works in th plane between the object and the surface when the object is not moving .\n\nStatic friction has values from zero to the value of limiting friction force \n\nExample if it is needed 40 N force to over come frictional force .\n\nif your pulling the object at 10 N force then the object will not move as it is smaller then the force needed to move it \n\nsince your your pulling it at 10 N force and the object is not moving , then Static friction force has a value of 10 N \n\nif you increase the force form 10N to some 35N , the object will not move at this time since it is smaller then 40 N . at this point the value of static friction will be 35N \n\nStatic frictional force is equal to the force at which your pulling the object and is developed in the surface of contact and opposote to the pulling force \n\nif you pull it at 40 N then the object will have just the motivation to move and just enough to over come the limit of static friction which is called the limiting friction .\n\nthank you .. i hope it helps you
961750
They specified a chain here for a reason.  Chains cannot sustain a compressive force.  The force on the table is produced only by the latest link to strike the table.  The part of the chain that has not yet hit the table is in freefall.\n\nSince they are asking for force over time, this really sounds like a momentum transfer problem.  Momentum has units like newton-seconds (force x time).  The force is then the momentum per second tansferred to the table.  Of course, the momentum is also mass x velocity.\n\nWhen the distance fallen is x, the link striking the table has fallen a distance of x.  It has a velocity based on the usual freefal relation:\n\n           v^2 = 2gx \n\nCall the total chain mass M, the length L and there are n links.  The mass and length of a link are M/n and L/n.\n\nThe momentum transferred by a link is P =  (M/n) * v \n\nThe time it takes from one link to hit until the next one is the link length divided by velocity:\n\n           t = (L/n)/v\n\nThe momentum is the product of force and time:\n\n P = F*t   so F= P/t\n\nSubstituting from above: \n\nF = ((M/n) * v) / (L/n)/v) = (M*v^2)/L\n\nSubstuting in the realtion between fall distance and velocity:\n\nF = 2Mgx/L\n\nAfter reading subsequent  answers, I notice that I completely forgot about the static potion of the problem, that is the dead weight of the chain on the table.  Since x length is on the table, that amounts to Mgx/L, which is exactly one half the static load.  Adding the two gives:\n\nF = 3Mgx/L\n\nYou can also work this via energy methods where you assume an elastic but stiff table.  The result is the same.
1171292
Sure, under non-equilibrium conditions.  Remember the Maxwell-Bolzmann energy distribution is only fixed at equilibrium...and thus the concept of "temperature" is only valid under equilibrium conditions.\n\nHowever, under pseudo-equilibrium conditions you can see a variance.  You may get one "temperature" for translational energy, and different (higher) "temperatures" for rotational and vibrational energies.\n\nYou can see this in supersonic expansions of gases into a vacuum.  The collision rate drops so rapidly that the higher "temperatures" of the rotational and vibrational energies are frozen in...no collisions are available to establish a true equilibrium w/ the translational energies.
390283
coefficient of friction and specific gravity
897524
Kinetic energy is .5 * mass * velocity ^ 2\nSo the change in kinetic energy, or delta K.E, would be:\n\n.5 * mass * initial-velocity ^ 2 - .5 * mass * final-velocity ^ 2\n\nSince Kinetic Energy is lost, the initial velocity goes first.\n\nGood luck on your test!
32662
actually the limiting friction is greater than kinetic friction.  it takes more force to overcome the forces of the limiting friction(start the object moving) then it does to overcome kinetic friction(keep the object moving). this makes sense logically, once the object is sliding, it's easier to keep it sliding then it is to start it moving in the first place...