1kg mass suspended from each side of a sting then wats the tension in the string?
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
133989
probably 2-5 grams.  it depends on the dimensions of the ballon.  you might as well ask a question like "how long is a piece of string?"
401877
Well, I can't even imagine what you are trying to let us picture but, at the same time, I feel this question must have at least one answer so, here I go without having a clear picture in my mind and perhaps lesser knowledge than required to answer this question-\n\nLets first imagine that this fictitious rope is made up of x number of neutrons and y number of protons.\n\nNow, Mass of x number of neutrons = "x" multiplied by 1.00864 atomic mass units.\n\nAnd mass of y number of protons = "y" multiplied by 1.00727 atomic mass units.\n\nNow, total "expected" mass of the rope = "x" multiplied by 1.00864 atomic mass units + "y" multiplied by 1.00727 atomic mass units\n\nTo Convert total "expected" mass of the rope from a.m.u. to kilograms :-\nUse the relation- 1 a.m.u. = 1.66 * 10 raised to the power (-27) kg.\n\nNow, weighing the rope, we will find that the actual mass of the rope is lower than the expected mass. Let the diference between the Actual mass and the expected mass be [M], now as Einstein told us:\nE = mc^2\n\nWe can express this mass difference as:\nBinding energy = [M] c^2\n\nI think that to determine the physical strength of this rope, one shall have to divide the Binding energy obtained above by the total lengh of the rope.\n\nAfter doing so, you'll get the value of the physical force that is keeping the rope intact.
909402
Well, it's Wednesday morning.  I hope this answer still has some value to you.\n\nI started by summing the torques at the hinge.  There are three:\nT(light)=5*9.8*7.5*cos(37)\nT(cable)=-tension*6.31*sin(37)\nT(pole)=8*97.5/2)*cos(37)\n\nThese all sum to zero.\n\nI computed tension=140.5\n\nThe sum of the horizontal forces must be zero.\nThese are the tension in the cable and the horizontal reaction force at the hing.  So the horizontal component is 140.5\nThe vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge\nthe vertical component is (8+5)*9.8\n=127.4\n\nI checked the answer by computing the magnitude of the reaction force two ways.  Using Pythagorean theorem \nsqrt(140.5^2+127.4^2)\n=189\n\nI then summed the forces parallel to the pole:\n5*9.8*sin(37)\n8*9.8*sin(37)\n140.5*cos(37)\n=189\n\nj
427198
I think I'm gonna look like I've got no clue whatsoever if I answer this one but, still, I'll give it a try risking my reputation. \n\nFirst of all, I didn't have a clue what you meant by a "buckyball", then I looked it up in a dictionary, now I know its a nickname for the C-60 Buckminsterfullerene. Thanks for adding one more word in my vocubulary (I desperately need a better vocubulary). \n\nNow, coming to your question, to determine the strengh of a rope made of C-60, umm.. I'll take a coward's way out by not going into the molecular constitution of the rope. As the physical strengh is an observable physical quantity, I'll solve the problem on a macroscopic scale. \nI'll hang a heavy mass with the help of the Carbon rope and will keep increasing the mass untill the rope ruptures.\nNow, I'll note down -\n1- The Weight (mass * acceleration due to gravity) that broke the rope.\n&\n2- The cross sectional area of the rope.\n\nNow, I'll just divide the Ultimate Weight by the cross sectional area of the rope to get the TENSILE STRENGTH of the rope.\n\nNow, I'll adopt the same procedure with the steel cable.\n\nI think, that way, we can easily compare the tensile strengths of the carbon & steel ropes.\n\nNow, to answer the second part, the density of the C-60 rope. Here too, I'll use a method that doesn't involve the molecular constitution of the rope (sorry again).-\nI'll use the Archimedes' Principle instead-\nLets first assign signs to the physical quantities-\nW = Weight of the Rope\nF = Buoyant force\nd1 = density of our rope\nd2 = density of the water (or the fluid in which you choose to immerse the rope)\nV1 = Volume of the rope\nV2 = Volume of water displaced by the rope.\ng = acceleration due to gravity\n\nTake the rope & put it into water (or any other fluid), now first see if it floats or gets submerged.\n\n\nIf it gets submerged, you can say that-\nW - F = (d1 - d2) * V1 * g \n(as the volume of the rope is equal to the volume of water displaced here)\nNow, knowing other quantites, we can calculate density of the rope!\n\nIf it floats-\nW = F\nd1 * V1 * g = d2 * V2 * g\n\nd1/d2 = V2/V1\n\nIf you take water as the fluid, d2 = 1000 kg/m^3
397084
Figure out what is was you did that caused the strain then do the same exact thing but in a backwards reverse motion! That should pull everything right back to where it started! If at first you feel nothing, then just remember second verse, not the same as the first! RICE!Rest ice compression and elevation! To elevate the groin, that is the question!
656278
tepends on the type of tension...
777446
If as described the entire meter stick is in balance, then the center of its mass/gravity would be where the support is, at the 22.4cm mark\n\naha ... here we go then,\n\n142.3 * (22.4-9)  = 134 * (x - 22.4)\n\nsolve this for x to get the answer\n( think of the entire stick as a ball placed at position x)
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
763604
a) The stress in the wire is given by force per unit area = m*g/A\nIf Y is the Young's modulus of the material of the wire, \nY = stress / strain. Therefore \nstrain = stress / Y.\nElongation = strain*original length.\nb) When the object is accelerated upward with an acceleration of a, the force acting in the wire = m*(g+a)\nThe rest of the equations above are valid.\nc) If the elastic limit of the wire is E, the greatest mass that can be accelerated up with an acceleration of a is got from \nE = Mmax*(g+a). Hence\nMmax = E/(g+a)
292220
Surface tension. The tape holds the balloon in place.
50723
its his weight * gravity. so the tension is: 75*9.8 which is about 735 N
368233
N is Newtons, a unit that measures force. For every 1kg of something, if it accelerates at 1 meter per second per second (m/s/s or meters per seconds squared), than there is a force of 1N acting on it.\n\nForce is calculated by multiplying the mass (not weight) with the acceleration. For example, if you are 75kg, than there is a force of 75kg X 9.8 m/s/s (acceleration due to gravity: roughly constant on Earth) = 735N pulling you down. BUT, there is a reaction force, in this case the NORMAL, pushing you back up at the same time and force. Therefore, the force pulling you down, 735N, take away the NORMAL, also 735N, and you will have a NET FORCE of 0N because you are stationary. \n\nSo in order to resist something pulling at 14N (ignoring air resistance, friction, etc), it needs to pull back 14N.
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.)
942786
The tension is the weight + the force Fb required to accelerate the bundle along with the rest of the elevator. \nNeed to calculate the elevator's acceleration a\ny = (1/2)*a*t^2 where y = 1m and t = 1.6s. Solve for a.\n\nNow calculate Fb\nFb = m*a. Solve for Fb.\n\nT = W + Fb
154867
The best tension is 1 or 2 above the max. lbs given to have the racquet strung.\nThe strings that do good for me are Ballistic Polymomo from Babolot. I break normal strings once a week, but with these it's about once a month.\nAll of this does the best for me.
109486
This is an Static problem:\nThe forces in the wire are:\nIn the middle of the rope the mass of Jonathan: 50kg.\nRigth side: T1\nLeft side: T2\n\nThis is a Vectorial equation: T1+T2=mg\n\nSince Johnatan is in the middle of the wire T1=T2... \n2*T1=mg\nT1 has two components T1x and T1y (horizontal and vertical)\nThe vertical component of T1 is the one that act against weight.\nT1y=T1*Sin(10)\nThen:\n2*T1*Sin(10)=mg\n[using mks system g=9.8m/s^2]\nT1=mg/[2*Sin(10)]\nT1=50*9.8/[2*Sin(10)]\nT1= 1410 Newtons.\nThe tension at "this point" is 2T1 =2820 N.\nHope it helps