Particles 1 and 2, each of mass m, are attached to the ends of a rigid massless rod of length...?
928343
Torque.\n\nThe weight of each particle is mg. The torque caused by particle 1 is mg * 20, counterclockwise, and the torque from particle 2 is mg * 80, clockwise. Therefore the net torque is mg * 60, clockwise.\n\nSo the force on the left particle is (mg * 60) / 20 = 3 mg, and the force on the right particle is (mg * 60) / 80 = 3/4 mg. Both of these are clockwise, so the left particle goes up and the right particle goes down. To get acceleration, just cancel m.
247359
Mass interacting with gravity
74552
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
1080952
mass, volume and temperature.
209689
mass in motion stays in motion unless disrupted by another force!
906099
T=2pi*sqrt(L/g) for pedulum with center of mass at the end, so rod length is twice that, so T=2pi*sqrt(L_rod/2/g), the answer is L=g*T^2/2/pi^2=0.243m.\n\nThe length is not 12.2cm because the center of mass of the rod is half distance of the length of the rod
936185
They have the same mass. An electron essentially has no mass. so both have the same mass.
904251
Let's say F1 and F2 are the tensions (forces) in the inner and outer sections of the rod respectively, w is the angular velocity, and r the distance to the center (half the length of the rod). Equating the radial force to (mass x centripetal acceleration) for each mass gives:\n\nr*w^2*m1 = F1 - F2 = 2F2\n(2r)*w^2*m2 = F2\n\nDividing the second equation by the first all variables cancel except for the ratio of the masses:\n\nm2/m1 = 1/4
777572
Centre of mass- is the point in or outside a rigid body along which(not necessarily at which) when a force is applied on the body, than the body behaves like a point mass, i.e. the body moves along the direction of applied force, without any rotational motion.\nConsider a system of particles of  a rigid body and choose a point of reference in the space. Let m1, m2, m3,…….be the mass of the particles and r1, r2, r3…………be their position vectors respectively, \nThen the position vector of the centre of mass will be \nR= (m1r1+m2r2+………..mnrn )/ (m1+m2+…..+mn)\nCentre of gravity- is the point at which the vector sum of the gravitational forces acting on all the particles of the rigid body acts, irrespective of the orientation of the rigid body. It can be found out by balancing the body on a pointer and equating the total clockwise moment and total anticlockwise moment acting on it.
609692
No. All motion is "Forward". Right down to the tiniest Particle. It can be pushed but not pulled. No matter the direction.
253742
Please take the time to\nvote for the best answer\n\nThank You!!\n\nRecall that an object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object: \n\nLook this up for more
405474
This is the rest energy of a particle. Energy equals mass times the speed of light in a vacuum squared. This expression became famous for its simplicity, even though it's not really all that practical. I read once that even at absolute zero particles will still have some motion. I'm not sure how one would go about proving that since it's impossible to get anything to a temperature of absolute zero. Anyway since there will always be some movement, the more practical version is:\n\nE = mc^2 /(sqrt(1 - v^2/c^2)), where v is the speed of the particle. (sqrt() is meant to mean the square root of the argument inside the parenthesis)\n\nIt's important to note that this is only true for a particle since you have to consider binding energies if you want to look at anything more complex.
605414
According to one theory there are particles, which are exchanged and travel between the objects, giving impression that objects attract or repel each other. Nothing is proved yet and it is just a theory, as far as my knowledge is concerned.
1095473
Corresponding to each kind of particle, there is an associated antiparticle with the same mass and opposite charges.
981397
This might be one of those questions that you have to answer with just using the formula, i.e. you will not get an actual answer, but an equation that if you knew the mass, length and force you could calculate the height dependant upon the force input, but thats pretty high level stuff, you would need so solve for angular acceleration and momentum, also in this sort of question you would be given either an impact force on a part of the rod (i.e. where the rod was hit) of you would be told the rod has some momentum. Then you could solve.\n\nI suggest you have another look at the question, I think you will find it says the rod is already moving.
874216
its newtons first law, (objects in rest stay at rest unless a force is acted upon it which causes acceleration) and objects in motion stays in motion\n\naka intertia keeps the smaller peices moving in the same direction
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!