A yo-yo shaped device mounted on a horizontal frictionless axis is?
813872
The applied torque is Fapp*R=70 Nm.\nThe tension on the rope supporting the 30 kg box is m*(g+0.8) = 318.2 N. \nThe torque producing this tension is 318.2*r = 63.64 Nm.\nThe remaining torque, 6.36 Nm, accelerates the wheel at 0.8/r = 4 rad/s^2.\nThe inertia of the wheel is torque/ang. accel. = 6.36/4 = 1.59 kg-m^2.
901129
(a) Assuming the merry go round is freely rotating (not powered by a motor), then it's subject to the conservation of angular momentum:\n\nI1 * w1 = I2 * w2\n\nIts moment of inertia is:\n\nI = I-disk + m*r²\n\nWhere I-disk is the given moment of inertia for the platform, m is the man's mass and r is the man's distance from the platform's axis of rotation. Thus:\n\nI1 = I-disk + (75 kg)(0m)² = 920 kg-m²\nI2 = I-disk + (75 kg)(3m)² = 1595 kg-m²\nw1 = 2 rad/s\n\nFrom the conservation of angular momentum, the final angular velocity is:\n\nw2 = (I1 / I2) * w1\n\nw2 = (920 / 1595) * (2 rad/s)\n\nw2 = 1.154 rad/s\n\n(b) The kinetic energy of a rotating body is simply:\n\nKE = (1/2) * I * w²\n\nSince we know the moment of inertia of the system for each case, along with its corresponding angular velocity, we can find the kinetic energy for both situations as:\n\nKE1 = (1/2)(920 kg-m²)(2 rad/s)²\nKE1 = 1840 J\n\nKE2 = (1/2)(1595 kg-m²)(1.154 rad/s)²\nKE2 = 1061.3 J\n\nFINAL ANSWERS:\n\n(a) w2 = 1.154 rad/s\n(b)\nKE1 = 1840 J\nKE2 = 1061.3 J
1142512
rotational sensor.. which is basically an inductive prox with a timming circuit to calculate revolutions.
254567
It is a horizontal belt which is mechanically driven in a continuous loop.  Items are laid on it and moved by the belt to eliminate manually carrying them.
64883
Up or away from the center of gravity.
109753
A mobius strip? or maybe Rotational Stability? It's an analysis of "wobbling", exhibited by common objects (coins, bottles, plates, etc) when they are spun on horizontal, flat surfaces. The object maintains a "wobbling" motion of a cylinder, which can be observed in slow motion by means of stroboscopic illumination.
139375
A horseshoe magnet allows you to concentrate more of the lines of force that eminate from the ends of a magnet through an object.  That translates in to greater sticking power.  \nYou can use a simalar trick with a bar magnet (like they do in hard disk drive magnets and the like) by applying metal to the rare earth magnets to do the same.
397135
The weight of the object is irrelevant unless one is taking into account air resistance. This is because the net force on the object is the force of gravity, so we have ma = -mg, or a = -g, which is all we need to know for the vertical kinematics equations. The only thing we need to know for the horizontal equation is the initial horizontal speed of the package.\nIf, on the other hand, one was concerned with air resistance, one would also need to know approximate air density and cross-sectional area of the object, as well as be able to numerically solve a nonlinear ordinary differential equation (the frictional force is proportional to some function of velocity) so I doubt this is what you wanted.
951551
A line that runs parallel to the Y axis.
964775
electropneumatic circuit \n\nhttp://www.festo-didactic.com/ov3/media/customers/1100/0000607001156324537.pdf\n\n:>peace
863293
In the opposite direction of the applied force or motion
343621
Try googling "magnetic levitation circuit"\n\nHere is one of the many hits I got.
609692
No. All motion is "Forward". Right down to the tiniest Particle. It can be pushed but not pulled. No matter the direction.
896989
You can make a ball bearing launcher:\nhttp://my.execpc.com/~rhoadley/maglnch.htm\n\nYou can make your own levitating MagLev train:\nhttp://my.execpc.com/~rhoadley/magtrain.htm\n\nOther info is available here:\nhttp://www.kn.sbc.com/wired/fil/pages/listmagnetsan1.html
149138
A machine that makes zogs...
110772
0.58 (with a horizontal bar right above the 58)
282311
a lever, a wheel, a pulley...