Really Hard Physics Problem..I Think?
966128
use the universal law of gravitation to find the force of gravity at the inner radius.  Use 1kg as your second mass (mass is irrelevent in this next part anyways), and use this force to find the accel of gravity.  Using the force and accel of gravity, you can find the kinetic energy needed in order to overcome the potential gravity of the planet (m cancels).  with kinetic energy you can find velocity, and with velocity and radius, period.  Hope that helps.\n\nfor the second part, repeat, except with the outer radius as D.
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
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.
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
833368
find the force of her weight. \nthat should be the answer
806553
C'mon, kid.  Go and read your book and do your homework.  I can understand if you are having problems with one problem, but you are posting this so that others will do your learning for you.  This is simple Newtonian physics.  Three laws you have to work with.  Basic, basic concepts.  If you can't handle this, then might as well give up your dreams of becoming a rocket scientist where you have to deal with special relativities and zero-gravity physics.  \n\nNow get moving!
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
757274
READ THIS CAREFULLY\nAs the block just slides on the table the net force just equals the limiting friction acting on the body\n\nhence,\nF=coeff. of friction * Nor,mal rxn\n\nnormal rxn= weight of body on a plane\n\nhence F= 20*9.8*0.4\n              =78.4N\n\nACC \nit is just begg to slide and the given data is just unsufficient 2 find the acc. of the box
50475
A)come on, the weight of the man is just his mass times gravity.\nB)this sounds like a torque problem, you need to know where on the beam the man is standing... since the beam is in equilibrium, you know that the sum of all torque is 0. 9im going to use T for torque because i cant get the greek letter i usually use) T=Fr (force times the distance from the pivit point) so all torques directed up minus all torques directed down equal 0. so that means all torques pointing up equal all torques pointing down. ok, so that means that:\n Tbeam + Tman = Tcable.\n and the cable is at an angle so it would actually be FrsinX instead of just Fr.  so now since the beam is uniform, we say that it's weight is in the center, so r is half of the length of the beam. so\n Tbeam=Fr=(288)(2.6) and now we convert the mass of the man to newtons, just multiplying by gravity(which im rounding to 10 because i dont feel like getting a calculator) and getting 750 Newtons. so that means\n Tman=Fr=750r. this has to stay as r because you didnt tell me where on the beam the man is standing. it probably indicate's where he's standing on your diagram. r would be the distance between the wall and the man because we're making the wall our pivot point, because if the end with the cable was our pivot point, everything there(including the cable) would be 0 because it would be 0m away from the pivot point, so we couldnt solve for Ft... so you'll have to plug in the "r" for the man yourself.  and Tcable=FrsinX=F(5.2)sin67 where F is the force of tension which is what you are solving for overall, r is 5.2 because the cable is 5.2m (the length of the beam) away from the pivot point. so now, we plug into what i told you before(Tbeam + Tman = Tcable.) which is true because the beam is in equilibrium. so \n\nTbeam + Tman = Tcable\nFr    + Fr   = FrsinX\n(288)(2.6) + (750)(r) = F(5.20)sin67\n\nplugging in the r which should be on your diagram, you can solve for F whichi is the tension in the cable.
760426
Figure out the centripetal force.  After that, this becomes a sliding problem turned on its side.
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.
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
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
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.
533021
there are some solved physics problems on \nhttp://www.dctech.com/physics/help/\nbut I don't know if it will help you out.
239996
Work done = force x distance moved by object i thinkin a straight line ,so ithink some conversion maybe needed.
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?