Physics Problem...Please Help.?
831197
Equation:\n\nV^2 = U^2 + 2 A (delX)\n\nThen solve F = M A\n\nwhere V = Final Velocity\nU = Initial Velocity\nA = Acceleration (unknown)\ndelX = distance of applied force (delX meaning delta X, or change in distance)\nF = Force\nM = Mass\n\nOf course you have to consider two things:\nThe force applied over the two feet would have to be constant (this is an assumption you have to make to solve)\nAll your units are in ft and weight.  you have to use the meters measurements and convert your ball's weight of 0.25 lb to kilograms (that's the mass measurement)\n\nyou now have more than enough information to solve this problem on your own.\n\nNOTE: Jim, that is a horrible misuse of the wrong equations for solving this kind of problem.  Nice try though.
188376
I was a rock star in college physics!   Top of the class!  But I can't remember any of it now.  My teacher always said to start each solution by drawing a picture, then making a list of the varriable you know, and writing down the equations you thik you might need to solve.  It worked for me!\n\nPost your problem, maybe I can help.  Traditional physics problems have not changed much in the last 5 years, maybe I can remember.
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
771534
The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
858739
Use Bernouillis equation and you will have your answer. You can use the equivalent lengths method to find the friction.
954443
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
1014316
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
981852
You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
948754
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
620382
Horizontal momentum is conserved throughout the problem. i.e., when half the mass fell downwards, all of the horizontal momentum must be carried in the second.\nThere's too much missing information for me to go farther than that...
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
153350
This problem employs the simple concept of conservation of energy (athough there are a number of ways to solve it).\n\nBasically, the person in the wheel chair has an initial kinetic energy at the bottom of thr ramp,\nKE = 1/2 mv^2\nAt the top of the ramp, some of that kinetic energy has been transfered into Gravitational Potential Energy,\nPE = mgh\n\nSince the mass of the person is constant, that drops out of the equation.\n\nSolve for the change in PE and subtract this from the intial KE.  Using the new KE, solve for the final velocity.
797830
Keep in mind these equations\n\nx(t) = x-initial + (v-initial)t + (1/2)(a-gravity)t^2\nv(t) = v-initial + (a-gravity)t\n\n(v-final)^2 = (v-initial)^2 + 2g(x-final - x-initial)\n\na-gravity = -9.8 m/s = -32 ft/s (i think..)\n\nTherefore\n\nx(t) = 0 + (64)t - (16)t^2\nx(t) = -16t^2 + 64t\n\n(v-final)^2 = (v-initial)^2 + 2g(0)\n(v-final)^2 = (v-initial)^2\n\nsince it is going in the opposite direction,\nv-final = -(v-initial)
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
833368
find the force of her weight. \nthat should be the answer