HARD Physics question 10 points to however gets the best answer?
959895
Man, that is a HARD one. You should give 20 points to the best answer.\n\n\nLong Live Jambi
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
843756
USE PRINCIPLE OF CONSERVATION OF linear MOMENTUM.\n\nearth gas lower KE because it has huge mass and huge inertia
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.
224170
please (2 pdf documents)\nhttp://physics.angelo.edu/~dbixler/phys3444lab4new.pdf\nhttp://physics.angelo.edu/~dbixler/phys3444lab8.pdf
905049
First assume that we neglecting any losses from air resistance or internal friction.\n\nWith that simplification, imagine two states:\n1) you are on the bridge 100m above the water and at rest\n2) you are suspended by the spring at its point of maximum extension, and therefore with zero velocity\n\nSince we assumed that there were no losses, the energy of both of these states is equal. Where is that energy located?\n -->1: your gravitational potential energy (no kinetic energy, no stored elastic energy)\n--> 2: your remaining gravitational potential energy, and the stored elastic energy in the spring (by definition, no kinetic energy)\n\nSo what has happened is that the spring has stored elastic potential energy that is equal to your loss of gravitiational potential energy. That's the only way the energies of the two states are the same.\n\nNow you can set up some equations. Let's say that you start at height H0 and end up at height H1.\n\nchange in PEg= m * g * (H1-H0)\nchange in PEk = 1/2 * k * [(H1-H0)-30m]^2\nwhere we have accounted for the 30m long unstretched length of the spring.\n\nNow by observing that -PEg = PEk, and plugging in the given values for m and k and the known value for g, you can readily solve for the change in height (H1-H0) (it should be < 0) It's a quadratic, so you expect to find one answer that is not physically realizable - reject that one, and select the other one.\n\nDoes that help?
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
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a) when it it closest to the gravitational source\nb) when the potential energy is minimum (same as a)\nc) speed = |velocity|\nd) momentum(p)= mv\ne) KE = (mv^2)/2\nf) usualy the potential energy is defined as 0 at infitity\ng) conservation of energy\nh) when it it closest to the gravitational source\ni) conservation of angular momentum\n\nThat should be enouph to figure out the answers... but think about it and draw some pictures so that you know it is right.
463641
see\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/surten2.html#c1\n\n here you can find the answer for your question with MATHEMATICAL EXPLANATIONS.
1006189
Explain Newtons laws of motion.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
1075431
Strength of material, velocity, distance; any of these ring a bell!
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.
641452
opposite force being applied
146531
This site is constructed well and has a high level of detail (up to degree level).\n\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html