what factors affect the spring constant of a spring?
80796
all physical characteristics of the spring contribute to the K or spring factor.
632008
Gravity, its weight, friction\nThat's about it.
731116
Spring constant: k = 4π²m/T² = 4π²(.712)/(2*.685)² = 14.976 N/m\n\nω = √(k/m) = 4.58625 rad/sec\n\nVmax = Aω = .102*4.58625 = .4678 m/s
971455
there are various factors which affects rate of reaction. like temperature,pressure and concentration of the reactants involved. and they all have different effects
972521
a. The spring constant for the inner spring can be determined by the fact that the force it exerts goes from 0 to 20 units (I can't read what the units are. Are they Newtons?) over a difference of 0.10 m. So its spring constant is 200 force units per meter, or 2 force units per centimeter.\n\nb. From point A to point B, the average force exerted by the spring is 10 units. Because the force changes linearly over the distance of 0.10 m, we can use this average and calculate the work done on the 5 kilogram object as 10 x 0.10 = 1.0 Force unit-meters.\n\nc. From point B to point C, the average force is 40 units, and the distance is 0.05 m, so the work done is 40 x 0.05 = 2.0 Force unit-meters. (Again, the force changes linearly over the 0.05 meters.\n\nd. The initial velocity of the object will be such that its kinetic energy equals the amount of work done by the two springs to slow its velocity to zero (i.e., to absorb all of its kinetic energy).\nJust use the formula for kinetic energy and set it equal to 3.0 force unit-meters.\n\ne. To do this step, you have to recognize that the inner spring exerts a force of 20 units at point B, and (assuming its spring constant does not change) it will exert a force of 30 units at point C. So the outer spring, which provides the rest of the force, is providing 0 units at point B and 30 units at point C, a change of 30 units in 0.05 meters, or 600 units per meter (= 6 units per centimeter).
883480
this seems an easy question but most of the student get it wrong since it is so easy to mess up your mind. it was a part of an exercise for my University exams. it makes you realise what is going on with spring ...just consider it as a opposing force. the body will drop and will expirience g --> it will accelerate. at the same time it will also expirience the sping;s opposing force.\n\nit will accelerate till the force become equal. this is the importnat factor. weight will keep on accelerating the body till it is completely cancelled by the spring force.\n\nso when \nFs = Weight then Velocity is maximum. \nso kx =mg =>\n\nx = mg / k.  (at this displacement the velocity  will be maximum)\n\nat this specigic point due to the theorem of conservation of energy we get that.\n\nkinetic energy + sping stored energy = potential energy at the start.\n\nso \n1/2mu^2 = (m*g)^2 / k => (be carefull mass = 76.7/1000kg)\nU = \n\nU = sqrt (( m*g^2)/k)) => u = 0.361m/s 9\n\nIF THE ALGBRA IS CORRECT\n\nNOW AFTER THAT POINT the body will still travel due to inertia and the springs force  larger now than gravity wll slow it down and bring it at rest \nat that point all the potential energy (due to height difference) will be transfornmed to springs stored energy therefore,\n\n1/2 k xmax^2 = mgXmax =>\n\nXmax = 2*m*g / k =>\n\nXmax = 0.02658m \n\n\n( i am at work and i believe the algebra is correct. the main point is to realise where certain things happen.)
454463
There are many answers to your question. Gravity obviously. Another example is if you are try to lift a car with a lever or using a pulley to lift something.
769961
Hi, \n\nIn general, the equation describing the position of an object under the influence of a constant force is as follows:\n\nh(t) = at^2 + vot + xo \n\nwhere : \nh(t) = height at any time t\na = acceleration\nvo = initial velocity\nxo = initial height.\n\nSo the answer would be \nF= m*a (mass times accelleration) = (the mass kg)*(1.96m/s^2) \n\nI'm guessing that you want to find out the equation for acceleration, so what you should do is use the values for t, and   h(t) to solve for the constants (xo, vo)\n\nexample: \nat time t=0, height is 0.50m\n\nso h(0) = a(0)^2 + vo (0) + xo = xo\n\nbut that also equals 0.50m;\n\nso:\nxo = 0.50m.\n\nand the equation becomes \nh(t) = at^2 + vot + 0.50m. \n\nThen use the different values of t, and h(t) to do the same thing with the other constant, and you'll have it. \n\nHope that helps, \nMatt
317039
Imagine the stretching of the surface of the drum as the stretching of a spring. use hooke's law to do calculation. u will need to know about the strength of material and etc. very very very complicated
522443
I hope, you mean frequency for pitch. \n\nThe frequency “n” is given by the formula, n = (1/2pi) x root of (k/m)\n\nk is the coefficient of stiffness of the spring and m is the mass attached to it.\n\nBy measuring the extension of length of the spring for a given force, the force per unit length gives the value of k.
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
963774
Use the fact that PV/T = a constant, we find the constant by pluging in the values we have, T = 20C, but we must convert that to K first so 20C = (273 + 20) = 293K.  And 1 atm should be converted to physicall units of Nt/m^2 for an easier solution.  1 Atm = 101325 Nt/m^2.    So our constant is (101325)(4.5)/293 = 1.336 x 10^7 so now you have one equation as\n\nPV/T = 1.336 x 10^7 setting T = 260C or 533K gives\n1)  PV = 7.121 x 10^10\n\nThe next equation that we need is the force on the cylinder end given the pressure, which would be\n\n2)  .01P = F\n\nThe spring will stretch a distance 3)  d = F / (2.00 x 10^3)  \n\nThe volume changes by  4)  V = 4.5 + d * .010   \n\nThis gives you 4 equations which you can now solve to find the solution for V, d, P and F
245329
Many of the manufacturers websites will have some background theory on their products. Here's an interesting website that claims to even anticipate instead of react to changing cable tension\n\nhttp://www.dfdinstruments.co.uk/ProductTechnology/towingsystem.htm
873474
these are some factors that i can thought of:\n1) addition/ removal of reactants/products\n2)increase/decrease of pressure of enclosed system containing only gases\n3)increase/decrease of temperature of enclosed system(only for reaction having enthalpy change)
454576
Because the weight applies a force to the spring.  F=-kx, so x = -F/k.  The great the force, the greater the stretch.  This is because many metals have memory and try to restore themselves to previous states.  If you bend it, the metal will try to bend back to where it was.
476332
A spring can be made of many different materials, it is usually of higher carbon steels heated and then quenched at a proper rate to give you the spring constant that you want.
564626
Which depends on the stres acted on the load and material used