howcanIgetawheeloffortunespinnumber?
343421
Well, you first off, have to know the english language  (spacing and punctuation included) then get on the show and when it's your turn you spin.
1069170
13212211\n\nJust counting numers in the previous number.\n\n'm i right??
791539
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
1028369
do a reverse number look up
195389
Rotational speed sensors have different types according to the operating prenciples of the design such as hall effect sensors, inductive sensors and magnetoresistive. The following link is about the hall-effect theory...\n\nhttp://w5ac.tamu.edu/~kc5nnw/359/wheelspeed/sensors.html\n\nThis is about a magnetoresistive type manufactured by philips:\nhttp://www.semiconductors.philips.com/acrobat_download/applicationnotes/KMI16-15.pdf#search='rotational%20speed%20sensors'\n\nhope I helped...
901198
answer is 3s. any no. multiply by 1 is the no. itself.
217405
You mean : Vehicle identification number?
519345
So you need to know the diameter of the rotating thing. Then it is simple dimensional analysis. Let's say the wheel rotating is 1m in diameter. So you have:\n\nR/M  and you want   m/s\n\nOk M->s is easy, you multiply by 60 (or in this case 1/60):\n\nR/(60M) = R/s\n\nTo change R to m, you need to find the circumfrance of the circle in meters. Which is pi * the diameter. So 3.141 * 1 = 3.141 meters. So each rev is 3.141 meters so multiply by that.\n\n3.141R/(60M) = m/s\n\nGood. So now you have a formula:\n\n[(3.141 * d)/60]*(R/M) = m/s
1094656
That all depends on the diameter of the wheels, and not on the speed of the car.  \n\nThe circumference of a wheel is given by c = 2(pi)r, which is 2 times pi times the radius.  \n\nAll you have to do is find out how many times the circumference of the wheel goes into one mile, which is given by,\n\n1 mile divided by 2(pi)r\n\nFor example, let's say the radius of the tire is 3 feet, and there are 5,280 feet in a mile, so you have:\n\n5280/(2(pi)3) = 280 (approximately)\n\nBecause the radius of the tire is in the denominator of the equation we just used, you can say there is an inverse relationship between the number of revolutions and the radius.  This means, as the radius of the tire increases, the number of revolutions made decreases.  \n\nIn fact, you could take it to the extreme, where the radius of the tire was about 840.34 feet, which would make the circumference of the tire equal to one mile.  In that instance, the revolutions would be equal to one.
838613
The Wc = M/L where M = the torque being applied to the spinning wheel and L = the momentum of the wheel.\n\nThis shows that if the wheel  spins more rapidly, a given torque on it will cause a slower precession (Wc).
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
725445
You can't without the radius of the wheel...m/s is a tangential speed.
967975
2A11\n4D13\n12G17\n48J25\n240M41
128153
You need to get the casting numbers off the engine.\nThis will show you where to find them\nhttp://www.mortec.com/location.htm\n\nAnd this will show what year and size it is\nhttp://www.mortec.com/castnum.htm
37325
first thing you need to do is calculate the moment of inertia of the wheel.  this can be done by treating the poles and the main wheel separately.  the moment of inertia of a circular disc (the wheel) is (1/2)MR^2, and of the rods (also a circular disc!) is (1/2)mr^2.  add them together to get the total moment of inertia.  \n\nyou'll need to use conservation of energy.  specifically, when the wheel drops a height h, the potential energy lost is -mgh.  this energy gets split into two parts, the rotational energy of the wheel and the translational energy (since it's moving downwards).  you can then write an equation setting the potential energy loss equal to the sum of the rotational and kinetic energies:\n\n-E_p = E_rot + E_trans\n-mgh = (1/2)Iw^2 + (1/2)(m + M)v^2\n\nhere I used (1/2)Iw^2 for the rotational kinetic energy and (1/2)mv^2 for the translational kinetic energy, where I substituted (m + M) for the total mass of your object.\n\nthe key point is now to realize that because the wheel unwinds along a string that is wrapped around the pole, the angular velocity w = v/r where r is the pole's radius and v is the translational velocity of the pole.  so you can substitute that into your expression.  it's not clear from your question whether you want the angular acceleration or the translational acceleration.  so you can either substitute the rw in for v, or v/r in for w.  \n\nin either case you then have a formula for the velocity as a function of the height of the wheel, which you can then solve for the velocity, and then take the derivative to get the acceleration.\n\ngood luck!
841789
Assume s=275, v=43, r=0.21 & t is time from start to stop; then v=a*t & s=a*t^2/2, ‘a’ being deceleration. Therefore t=v/a, s=a*(v/a)^2/2; thus 2*s*a=v^2 or a=v^2/(2*s).\nAngular deceleration w’=a/r.\nNow I=0.175, so frictional torque T=I*w’. plug numbers in.