Actually what is torque and speed?what is the relation between them?
968030
Torque = Force * Distance(from fulcrum)\ntherefore\nTorque =(mass * acceleration) * D\nThe higher the torque, the greater the acceleration, but you'll need a duration (time) or distance travelled to determine a velocity.\n\nMore torque = more acceleration.
614422
Depends what information you already have. The simplest one I can think of is:\n\nVelocity = Acceleration x Time\n\n-Therefore-\n\nAcceleration = Velocity / Time\n\nBut you might have other variables involved here.
779727
combustion? your question i super vague. the drive shaft applying torque created by the engine to the wheels which turn making the car move? you might want to be more specific. and if you look at wikipedia.org under combustion engine you might find what your looking for
876839
speed is velocity - both measured as the change of distance divided by the change in time. You may mean the angular velocity, how fast is it spinning; or how fast the air is moving around the center (relative to the air not spinning); or how fast the tornado is moving across the ground.
662957
Many people are often confused about power and energy.\nLet's start from the beginning, with some basic equations:\n\nEnergy = Force * distance\n\nFor a mass m moving at a speed v, the energy E is:\n\nE = 0.5 * m * v * v\n\nEnergy can be measured in Joules (J) or British Thermal units (BTU). 1 BTU is equal to roughly 1055 Joules. If a 2100 pound car is travelling at 60miles/hour, the energy is about 342600 Joules or 325 BTUs. This is the same amount of energy required to accelerate the car from 0 to 60 miles/hour on level pavement. This is independent of acceleration, it doesn't matter if it takes 6 seconds or 60 seconds, it's the same amount of energy.Energy is always conserved, meaning it can't created or destroyed, it can\nonly change form.\n\nPower = rate of energy\n\nPower can be measured in Watts (W) or Horsepower (HP). 1 HP is the same as 746 Watts.\n\nPower, unlike energy, is not always conserved. If we have a system where energy is flowing at a constant rate, then power is conserved, meaning the power in is equal to the power out. For standard cars, power from the engine is delivered to the wheels through a transmission. There are some losses due to the transmission, air conditioner, alternator, and other systems, but this is not considered a significant loss. If the air conditioner is turned off, your headlights are off, and you have a manual\ntransmission, the losses will be not more than a few percent. Automatic transmissions only increase losses.\n\nThe gear ratio is a transmission does not change power output. Remember,power is conserved in this case. There might be some slight change in power loss for different gear ratios, but for the most part, it's insignificant.\n\nSince we know power is the rate of energy, we can divide the energy calculated above by the time required to produce the energy. If it takes your car 6 seconds to go from 0 to 60 miles/hour, then the power is about 57000 Watts or 76.6 HP. Keep in mind, this is the average power, it doesn't\ntell us the power at any time during the acceleration. We would need detailed speed measurements from start to finish to calculate that.\n\nGiven these numbers, I would guess the engine is capable of producing 150 HP or more. How did I come up with that number? Internal combustion engines, like gasoline, Diesel, and even turbines, do not have a flat power curve. The power output is zero at zero RPM and increases more or less\nlinearly until some peak. In stock gasoline engines, the peak power is at 5000-6000 RPM or so. As a rule of thumb, the peak is generally 1000 RPM below red line. So if you could run your engine at 1000 RPM below red line,\nthen you could achieve near peak performance. If watch your tachometer, I would guess that the RPM averages much less than that, maybe even half.\nThat's why you are getting only half of the power the engine is capable of producing.\n\nRPM, power and torque are all related. Torque is simply power divided by RPM. The power out of a transmission is the same as the power in (minus small losses), but the RPM is different depending on the gear ratio. If the output shaft spins at half the speed of the input, then the output torque is double that of the input. RPM doesn't really matter in the end as far as power is concerned, remember, power is conserved. Changing gear ratios does not change the power, it only affects torque and RPM. It's easier to start with calculate power and work out RPM and torque if you need it. You can do it the other way, but it's easy to make mistakes.\n\nThe reason most cars have a transmission at all is because gasoline engines cannot produce torque over a wide range of RPM, particularly at zero RPM.Diesel electric locomotives and subways do not have transmissions with variable gear ratios because they use electric traction motors instead of\ntransmissions. Electric motors can produce torque over a wide range, even at zero RPM. This is known as the stall torque
834512
Torque is not the right question to ask in this problem!\nFirst define the generator output in terms of watts.\nAssume an efficiency of 0.8 for the generator and define the Input of the generator (In = Out / 0.8).\nThis is the output of your motor (assuming some diesel/petrol).\nTorque is expressed as N/m (force x radius) and is part of the electric generator specs.  See there to get the equivalence.\nUsually, the motor power is also 0.8 efficient, so the motor power should be W / (0.8*0.8) and that will give you the right ratio.
448583
what do you mean?\nhow to calculate their speed or wat?
144016
The net force needed is zero.  Force is mass * acceleration, and acceleration suggests a change in velocity.  For constant speed and direction, the acceleration is zero, and hence the net force is zero.  When moving at constant velocity, the frictional and drag forces (which work to slow the car) and the drive force of the engine are exactly equal and opposite.  The net force is the sum of these two opposite forces, and hence the net force is Zero.
142488
Instantaneous speed, as the name indicates, is measured at a particular instant of time. From a math standpoint, by calculating the first derivative of the displacement function, you can use the result to calculate the speed at any instant of time. In other words, the instantaneous slope of the displacement function is equal to the instantaneous speed.\n\nIf you look at graph of a moving object (displacement vs. time), if you see a steep slope, this means that the object's speed is fast relative to a curve with a moment of shallow slope. How much distance are you moving per unit of time? The more distance you move, the faster your speed. Decrease that unit of time to almost 0, and you'll get instantaneous speed.
852750
T=I*w’, T=16 torque, w’= angular acceleration, I=1.2/1000; In time t (SECONDS) its angular speed w=w’*t;\nOr w’=w/t or w’=T/I; thus w/t=T/I or I*w=T*t is angular momentum, & w=T*t/I
53841
The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
42744
momentum is speed and mass is the substance of objects
890959
I think I know what is confusing you. Total angular momentum of your system M=M1+M2=I1*w1+I2*w2, while I1=I2, w2=-w1, Thus M=I*0=0. Angular momentum is a vector, because w (angular speed) is a vector. How could we describe the difference of the system with w=0 and w not equal to zero? If w=0, then kinetic energy of the system E=0. When they are rotating, kinetic energy of the system E=E1+E2=I1*w1^2/2+I2*w2^2/2, which is not 0, as both w’s are squared. That means you have to do some work to obtain a certain angular speed w no matter what direction disks would spin.
26747
ok.  call the downward translational velocity of the whole thing v.  this means that the tangential velocity at the edge of the poles (at the radius r) is equal to v.  the reason for this is that the strings are wound AROUND THE POLES, so this winding and the condition that the string has to unwrap for the thing to move, sets the angular velocity w = v/r.\n\nnow, if you look at the main wheel part, the angular velocity has to be equal to the angular velocity of the poles, since they are one solid piece.  however, as you note the radius of the wheel R > r.  so, the fact that v / r = w = V / R implies that V > v.  in other words, the motion of the outer edges of the wheel is faster than the motion of the outer edges of the rods.\n\ni think you are getting confused by saying that V is the translational velocity of the wheel.  it's not.  v is the translational velocity of the wheel, and of the rods.  the fact that the string is wound *around the rods* and not around the wheel sets the angular velocity of the whole apparatus equal to v/r.  V, the tangential velocity of the outer edges of the wheel, is much faster than the translational velocity.\n\nthat clear?
193077
Force=Mass x Acceleration\n\nSo Acceleration = Force/Mass\n\nHope this helps
838774
each person produces a torque of:\n\n4.1m x 180 = 738Nm\n\nSo total torque is 2 x 738.
131445
Torque is force*distance.\n\nFind the weight of the person (M*g) = 89kg *9.81m/s^2 \nM*g= 873.09 Newtons\n\nthere is a 1:3 lever arm there--> 4cm/12cm\nThe muscle is having to exert a force of 3*873.09N or 2619N\n\nThe torque around the pivot is 2619N @ 4cm (or .04m)\n2619N * .04m = 104.77N*m\n\nYour question says you want to "find the torque" yet you ask for the "forces pushing up".  Make sure you know which one you need.