Hard Physics Problem!!!?
971439
First of all, the answer is approx. 1691 revolutions/day.\n\nTo get this answer, a few fundamental equations/concepts are needed. \n\nKnow that the inward acceleration felt by an object on the radius of the space station is equal to the radius of the space station times the angular velocity of the station squared:\n\na = r*w^2\n\nThe needed acceleration is ~9.81m/s^2 (1g). Using this and the the station's radius (650m), the angular velocity is 0.1229 rad/s.\n\nConvert this to revolutions per second knowing that 1 revolution = 2*pi radians.\n\n0.1229 rad/s divided by 2*pi rad/rev = 0.01957 rev/s\n\nMultiply this answer by 86400 s/day to get 1691 rev/day.\n\nGood luck, I hope this helps!\n\nPS - I used http://en.wikipedia.org/wiki/Circular_motion to confirm the formula used in the beginning.
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
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
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
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.
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.
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
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...
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
971746
This problem involves kinetic energy, force of kinetic friction, work, and conservation of energy.\n\nGiven: weight of girl + sled=645N\n           mass of girl +sled=645/g\n                                        =6.58N\n           speed of girl +sled at bottom=7m/s\n           coefficient of kinetic friction=0.050\n\nKinetic Energy, KE at the bottom of the hill=1/2mv^2\nKE=1/2*(6.58)7^2\n     =161.2 Joules\n\nForce of kinetic friction, f=uR, where u=coefficient of kinetic friction, and R is the force normal to the surface equal to weight of girl+sled.\n\nf=0.050*645\n =32.25N\n\nWork=f*s, where f is the force of kinetic friction and s is the distance the sled travels. Thus\n\nWork=32.25s\n\nBased on principle of conservation of energy:\n\nKE=Work\n161.2=32.25s\ns=161.2/32.25\n  =5.0m
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
746965
You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
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.
735831
Did you put your entire homework assignment on here? I noticed you posted a bunch of science questions just now.\nIf you don't have your textbook handy [you really should be looking up these answers so that you can remember and understand them when you have your test/exam], go to the links below and look them up. The 1st is a down loadable physics textbook. The 2nd has animations of physics.
754995
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5