how to get the speed,tension and weight.....physics help(formula)?
970605
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
72553
you find the force of gravity, but there is more force acting on the gymnist, so you must add that force too. the force of gravity is mass times gravity so 72 x 9.8 and the other force is mass times acceleration so thats 72 x .23. then you take these two forces and add them together to get the tension in the rope.
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
12996
I can't draw a free-body-diagram here, so it's going to be somewhat more difficult to explain (a picture is worth 1000 words and all)...\n\nSince the block is moving at a constant velocity, we know that all the Forces are balanced.  this means that the Force of friction is equal to the force of gravity pulling the block down the incline.\n\nI'll write this as \nFf = uN  \n\nSince we know the angle of the slope, 25 degrees, we can break the Force of gravity into 2 components,  one that points down the incline (i'll refer to it as Fx) and one points the opposite direction (I'll refer to is as Fy) of the Normal vector N.\n\nIf you draw that triangle you notice the the hypotenuse has magnitude(length) of mg (mass times the gravitational constant.\n\nnow, lets use the sin & cos relationships to determine the length of the 2 components: Fx & Fy\n\nsin = opposite / hypotenuse\ncos= adjacent / hypotenuse.\n\nIf you look at the diagram, the opposite is Fx so\nsin 25 = Fx / mg \nFx = sin25 * mg \n\nand the adjacent is Fy so\ncos 25 = Fy / mg\nFy = cos25 * mg  \n\nagain since all force are in balance, \nFy = N \n\nand Fx = uN \nsin25 * mg = u * Fy \nsin25 * mg = u * cos25 * mg \nwhich reduces to\n\nsin25 / cos25 = u \nand \nfinally\nu = tan25 = ~.466
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
149886
Hope this helps,  first remember that energy must be conserved so E initial will equal E final so you use the equations for potential and kinetic energy.  mgh for potential and 1/2mv^2 for kinetic.  You know the initial velocity, gravity and the height of the ramp so you simply break it into components and get the answer of 3.15 m/s.  HOpe this helps
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
677540
A lot of physics (especially beginning physics) is just finding the right equation and plugging in what numbers you are given.\nEquations usefull for your problem:\nKE = 1/2*m*v^2 (KE = kinetic energy, m = mass, v = velocity)\nAND\nWork = Force * distance     This is a very important equation\nit's easy to memorize so do it!
730564
1. friction\n2. W=mg  , he has a mass of 100Kg. Weight is the force=newtons, g=pull of gravity N=100 x 9.8m/sec^\n98Newtons\n3. correct
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
905915
You're going to have to resolve this two-dimensional problem into two one-dimensional problems.  I'd suggest using vertical and horizontal.  On the vertical side, we have gravity pulling down, and the normal force and one component of the tension from the rope pulling up.  This, by the way, is why the normal force is not equal to the weight - by pulling on the rope, you're partially lifting the sled off the snow.\n\nYou know that there is no acceleration vertically, so the net force vertically must be zero.  Knowing this, you can get the normal force as a function of the tension as the weight of the sled minus the vertical component of the tension.  Knowing the normal force and the coefficient of friction will give you the force necessary to pull the sled at constant velocity.  (If the velocity of the sled is constant, what can you say about the net force horizontally?)  Since the friction is only opposed by the horizontal component of the tension, you can find that.  Once you have that, you'll find that the only variable you have left is the tension in the rope, and you can solve for that.\n\nHope that helps.
75568
There are four (at least that I know of) to choose from.  You have to first analyze your problem and find out what you know and then solve for your unknown.  However, these equations assume that the acceleration is constant.\n\nv = v0 + a(t)\nHere:  v=final velocity; v0=initial velocity; a=constant acceleration; t=time\n\ns = s0 + (1/2)(v0 + v)(t)\nHere:  s=final position; s0=initial position; v0=initial velocity; v=final velocity; t=time\n\ns = s0 + v0(t) + (1/2)(a)(t^2)\n\nv^2 = v0^2 + 2*a(s-s0)\n\nI didn't say what each of the variables in the last two equations stood for because they are the same as what you see in the first two equations.  One good way to check if you are on the right track or not is to make sure your units work out right.  For example, if you are looking for velocity, chances are you'll be looking for something like meters per second (m/s) or miles per hour (mi/hr or mph).  Hope this helps!
217280
According to Newton's second law,\nF = m*a \nF = Force\nm = mass\na = acceleration\n\nnow, also,\n\na = (v-u)/t \nv = final velocity\nu = initial velocity\nt = time\n\nput this expression into first equation, you get\n\nF = m (v - u)/t\n= (mv - mu)/t\n\nalso, momentum = mass * velocity\n\nso,\nF = (final momentum - initial momentum)/t\n\nwe can also say that\n\nF = Rate of change of momentum
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.
844762
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you