Given magnetic field, particle charge, and force (vector), how can I calculate velocity's x and y components?
406322
q*(x,y,z)X(a,b,c) = q(yc - zb, az - xc, xb - ya). In your case, B is (0, 0, -1.2) so you have -5nC(-1.2y, 1.2x, 0) = (-3.6*10^(-7), 7.6*10^(-7), 0) which is a trivial system of equations.
959062
The ions are acceerated through the potential difference V. The kinetic energy they achieve going through that potential is\n\n0.5*m*v^2 = Q*V\n\nwhere m is the mass, v is the speed, Q is the charge, and V is the potential difference.\n\nThe moving charge in the applied magnetic field will result in a force that is at right angles to the motion. The ion will traverse a curved path, and as a result will experience centripetal acceleration. The force of the applied magnetic field (Q*v*B) is balanced by the centrifugal force.\n\nQ*v*B = m*v^2/r\n\nQ*B = m*v/r\n\nUsing v derived from the energy balance above\n\nv = sqrt(2*Q*V/m)\n\nso that\n\nQ*B = (m/r)*sqrt(2*Q*V/m) = sqrt(2*m*Q*V)/r\n\nSolving for the mass m\n\nm = (Q*B*r)^2/(2*Q*V)\n\nsimplifying\n\nm = (0.5*Q/V)*(B*r)^2
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
716372
1 and 2 are easy. I'll help with #3.\n\n3. You need I, the rotational inertia. From a table in my physics book, I = MR^2 for the reel. You need to convert the force to torque, T = F*R. \n\na. Then you can get the angular acceleration, alpha, from \nT = I*alpha\n\nb. In t = 0.25 s, omega will reach alpha*t. The angular displacement, theta, will be (1/2)*omega*t. From that and the circumference you can get the line drawn from the reel.
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
807345
Kinetic energy is m v^2 / 2.\nTo find v, you need to find the magnitude of the vector that represents the object's velocity. (Actually, the above expression requires the SQUARE of the velocity, which will save us a step in the solution.)\n\nUsing vector algebra and the Pythagorean theorem, the velocity is sqrt (6^2 + 2^2). Don't bother to take the square root and you'll have v^2, which you can use in the equation for kinetic energy.\n\nThe velocity changes by 8i+4j. Let's assume that this change occurs over a period of 1 second. (It turns out that the amount of work done does not depend on the time period, so this assumption doesn't affect the answer that we will calculate.)\n\nTo change the velocity by 8i+4j, we need to apply a force of 24 newtons (8 x 3) in the i direction, and 12 newtons (4 x 3) in the j direction for 1 second.\n\nThe amount of work done by each of these forces (considered separately) is equal to the force times the distance over which it acts. The 24 newton force in the i direction acts over a distance of 10 meters (since the average i-direction velocity during this 1-second period is 10 m/s; you calculate that by averaging the initial and final velocities, 6 m/s and (6+8) m/s). So the work done is 240 newton-meters, or 240 joules.\n\nIn the j direction, the force of 12 newtons acts over a distance of 1 meter, for total work done of 12 joules. (Note that the average velocity in the j direction is 1 meter/second, found by averaging -2 m/s and (-2+4) m/s.)\n\nSo the total work done is 252 joules. (I'm a little uncomfortable about treating these separately and adding them up. If I had a physics book here, I'd read it to decide whether this is the right treatment. If you have a physics book, I suggest you do that.)\n\nP.S. I have assumed that the velocity changes BY an amount of (8i+4j). If instead it changes TO (8i+4j), then you need to figure out the amount BY which the velocity has changed, and plus those i and j values in where I used 8 and 4.
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.
459989
According to the Lorentz Force law:\n\nForce=q[E + v x B]\n\nso if there's no Electric field it will just be q(v x B) where q is the charge of the particle
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
871196
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
840248
Vertical components\n\nv = u + at\n0 = u + (-32)(2.1s)\nu = 67.2ft/sec\n\nMaximum height the ball reached\n\nv^2 = u^2 +2gs\n0 = (67.2)^2 + 2(-32)(s)\n64s = 4515.84\ns = 70.56\n\ndownward velocity 0.8 seconds after the ball reached the max.height.\n\nv = u +at\nv = o +(32)(0.8)\nv = 25.6 ft/sec\n\ndistance from the maximum height\n\nv^2 = u^2 +2gs\n(25.6)^2 = 0 + 2(32)s\ns = 10.24 ft\n\nHeight of the fence = 70.56 ft - 10.24 ft\n\n\nHeight of the fence is 60.32 ft...\n\nthere ya go...
846846
Very easy.  If you know which direction the object is going, then you know the direction of your velocity vector - the angle (let us say "c") it makes with x axis.  Draw xy coordinates on your paper to represent a plane that contains your velocity vector.  You can draw a line parallel to x axis from the tip of the velocity vector.  That will give you the vertical velocity.\n\nIf the angle is "c", for example, your horizontal velocity is M cos c and your vertical velocity is M sin c, where "M" is the length of your velocity vector, that is it's numerical value.\n\nYou can also
476628
In physics, the Lorentz force is the force exerted on a charged particle in an electromagnetic field. The particle will experience a force due to electric field of qE, and due to the magnetic field qv × B. Combined they give the Lorentz force equation (or law):\n\n F= q(E + v x B)\nwhere\n\nF is the force (in newtons) \nE is the electric field (in volts per meter) \nB is the magnetic field (in webers per square meter, or equivalently, teslas) \nq is the electric charge of the particle (in coulombs) \nv is the instantaneous velocity of the particle (in meters per second) \nand × is the cross product. \nThus a positively charged particle will be accelerated in the same linear orientation as the E field, but will curve perpendicularly to the B field according to the right-hand rule.
1108157
You should be able to measure all three, in both cases.. limited only by the sensitivity and accuracy of your instruments.\n\nwhen a particle with mass moves past you, or you move past it, the force excerted by gravity between you and the particle changes with distance. Of course you need to have a mass too. and your instruments need to be good enough to distinguish these forces from other gravitational and other kind of forces.\n\nIf a charged particle moves, it produces a magnetic field. this is the same effect as seen in a elecromagned where moving elecrons produce magnetic fields.\n\nif a charged particle moves past you, and you have a detector that is based on charge too, you will notice an increase or decrease in attractive or repulsive force between your detector and the charged particle as the distance between the both changes. again this force is miniscule and is easily confused with other stray forces including the above gravitational ones.
247677
There's this equation called Coulomb's Law that relates charge and force: F=k*Q*Q/(r^2), where k is Coulomb's constant (8.988e9) and r is the distance between the two charges. That doesn't give us the units for an electric field, which is a force per charge. But we can divide both sides of Coulomb's Law by a Q, giving us F/q=k*Q/(r^2).\n\nTo find the magnitude and direction of the net electric field, you need to add together the electric field vectors created by each of the two charges. For the first charge, the electric field is k*15e-6/(.12^2)=9.36e6 N/C. The electric field of the second charge is k*10e-6/(.05^2)=3.60e7 N/C. Make sure to change cm to meters, and note that the charge q at which you are finding the electric field does not factor into our equation. Hopefully the problem describes in more detail the location of the two charges so you can determine the appropriate vector sum's magnitude and direction.