what are 2 forces needed for this question?
92221
Gravity and electromagnetic forces.  Gravity is an odvious answer.  Most other forces, including frictional forces, fall under electromagnetic forces.  The strong and weak force mentioned earlier can not be experienced directly.  They come into play when dealing with radiation and sub-atomic physics.
659614
In order to get better responses, you will need to more specific and explain what you mean by that "question".\n\nThe first thing that came to my mind was centripetal force which maintains circular motion.  Without centripetal force, the object would not be in circular motion (thus linear motion).
539746
F = ma\n\nF - Force\nm - mass\na - acceleration
714549
a)\nConsider the energy of the sled:\nIt loses potential energy of m*g*R*(1-cos(a))\na is the angle that has opened from vertical\n\nThe kinetic energy is\n1/2*m*v^2\nso v=sqrt(2*g*R*(1-cos(a)))\n\nfor b\nthe normal force is\nm*g*cos(a)\nthe centrifugal force is\nm*v^2/R\nv=sqrt(R*g*cos(a))\n\nfor c\nusing the maximum velocity for departure as\nv^2=R*g*cos(a)\nplug into\nv^2=2*g*R*(1-cos(a))\nR*g*cos(a)=2*g*R*(1-cos(a))\ncos(a)=2(1-cos(a))\ncos(a)=2/3\nabout 48 degrees\n\n\nj
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
250025
The net force is zero.  Remember the equation F=MA:  force equals mass times acceleration.  If something is at rest, A is zero, so F must be zero.\n\nThe gravitational force on the brick is exactly balanced by the force of the table holding the brick up.  Net force is zero.
772638
try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses
432088
*copied from previous post\n\nYour question is unclear about what set of assumptions you want to use. You seem to imply a perfectly constructed stool, with a perfectly centered force, using incompressible material. You want the stool to have no mass, and you imply that you want to use statics to solve this problem. Strictly applying statics, this system is indeterminate, NOT the ridiculous answer of 0 [N] in two legs and W/2 [N] in the others. There are only 3 equations of equilibrium here, 1 for vertical forces and 2 for moments. Since you have four legs, this system cannot be solved by statics alone. Treating the stool as a deformable structure, you can use geometric compatibility to get a fourth equation. I challenge you to find a competently built stool that has 0 [N] in any leg (a physical stool built of real materials), or an incompetently built stool with 0 [N] in any two legs (unstable system that is obviously not static). The answer indicated as "correct" requires an incorrect application of statics. A large part of studying physics is learning when to use a particular tool, and the physical limits of that tool.
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
829880
the equation is:\nF=G*m1*m2/(r*r);\nso, F=6.67*10E-11*0.14*0.17/(0.65*0.65);\n\nthis is the force that either will exert on the other
731589
Well To solve this problem, you need to break up forces in x and y components. Once you find both x and y components, you can find magnitude of force and direction ken must pull. \n\nTo keep the bear from moving, some of the forces (in both x and y direction) equal to zero. \n\nSo, sum of the forces in X-direction = 350sin(60) - 400sin(45) - F(in x direction) = 0\nSum of the forces in y-direction = 400cos(45) + 350cos(60) - F(in y direction) = 0\n\nSolve for F(in x direction) and F(in y direction), you will get these: F(in x direction) = 20.266 lb east and F(in y-direction) = 457.84 lb south\nTo find magnitude = ||F||= sqrt((F(x))^2+(F(y))^2) = 458.29 lb\n\nTo find angle= tan(theta) =(F(y))/(F(x)) where theta is the angle\ntheta = arctan (F(y)/F(x))= 87.5 degree east south\n\nSo Ken must pull the rop with force 458.29 lb in the direction East 87.5 degree South to keep the bear from moving. \nI think this is right. Good Luck.!!!
313833
So, this looks like a homework problem, but this is how you would set it up.  And, it looks like the beam is massless...\n\nFirst, you know that the beam's not going anywhere, so the sum of the forces on it must be equal to zero.  That is, the sum of the masses must be equal to the sum of the forces exerted by the two supports-- 5N+80N+30N=s1+s2, where s1, s2 are the two forces.\n\nNext, you know that the beam's not rotating, so the sum of the torques on it  around any point must be zero.  To make things simpler, I'll set up the equation with the torques about end A--  5N*1m+80N*2.5m+30N*1.25m-s2*4m=0.  Note the difference in sign of the torque produced by s2 relative to the other forces.\n\nNow you see that the second equation has only one unknown-- s2.  Solve for s2, then plug that value into the first equation.  Then solve for s1.
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
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
1036294
letter b is the answer.. because the weight of the object is approx. 627.20kg multiplied by 9.8(acceleration due to gravity) so the product is the force..