details of bearing in a car gear box according to load ?
482206
Is exponentially proportionally to the load.
1121436
let's see. The car in on large piston, right?\n\nso the pressure increases by 90/250 units.\n\nthe additional weight needed to compensate from from the small piston side is (90/250)*10=90/25 kg\n\nthis weight must be provided by a weight of oil going up in the small piston, it's volume being 90/25/560 m3, \n\nso the large piston must go down to provide this volume by\n90/25/560/.025=0.2571 m  (slightly under a foot)\n(m2 = 10,000cm2, so 250cm2=0.025m2)\n\nSo I guess 0.2571 m  is the answer. \nDo not remember any of this physics, but this seems to be the only way to link the numbers.\nLemme know if it is right :)
541033
Total weight the vehical or product is built for according to the mfg. specs.
919976
Apparently we are ignoring any pitching motion due to their getting in the car and due to the bump, since you'd need to know a lot more about mass distribution if it was pitching. So we're only considering vertical movement and the distance between axles is irrelevant. I'm assuming "a total of 8.9 cm" means the sum of the 4 spring deflections, so their average deflection is 2.225 cm. If it actually means the average deflection, then k and m are 1/4 what you'll get with the following calculations. Stiffness k (of the 4 springs in parallel) in n/M = 146*g/.02225. Natural radian frequency w = sqrt(k/m), so m = k/w^2 = k/(2*pi/1.7)^2. (That's the total m; mass of car = m-146.) \nThis was not a well-phrased question, because "a total of 8.9 cm" also could mean 8.9 cm from their no-load length (and thus due to the weight of the car and the students). However this would result in an insoluble problem. \nBTW, why are they geeky? Did they figure all this out before me?
939535
Sorry, but you'll need to give a little more information. How are the pulleys and weights arranged? Just a little picture would help.\n\nTypically, pulleys, weights, tensions and accelerations are extensions of the block on an incline problem except the pulleys are providing the force vector direction change.\n\nIt might help to think of the problem as a sum of vectors and look for where their vector sum lies.
1026758
friction is equal to the normal force (such as weight) times the coefficient of friction. f = c of friction X normal. if you try to push a cardboard box on the ground, its kinda easy. now imagine putting a hundred pounds in that box. its harder to push becuase there's more friction holding the box back.
437681
You have to calculate the centripetal force.\nFc = Mrw^2\n   = (M V^2)/r\n   = [100*(10m/s)^2]/30\n   = 10000/30 m/s^2\n   = 1000/3 N = 333.33 N\n\nM = Mass of the car\nV= velocity\nr = radius\n\nFor banking curve \ntan theta = v^2/g*r\ntherefore\ntheta = arctan [10 m/s]^2/[9.8 m/s^2  * 30m]\n      = arctan[10*10]/[9.8*30]\n      = arctan[100/294]\n      =    degrees (please chack your tables and get the arctan)
598306
nothing-\nthe system of atoms inside your car is stationary relative to the baloon.
391537
technical manual 1986 buick regal
220156
There is a gear, at the end of the cam shaft, that drives the distributor shaft. On the other end of the cam shaft, is a gear that is driven by a chain, or belt, that is driven by a gear on the crank shaft. They all work together to 'time' the ignition of the fuel/air mixture compressed in the cylinder. Read 'Chiltons', refer to your car's year and model.
670662
Try Global Spec. It's free and allows you to put in your desired torque and rpm, then shows you matching products. Keep in mind they will spam the email account you register with mercilessly.
221158
A good site is Weight Weenies.  Shows the weight of many bicycle parts.
336317
It is very difficult to accurately describe a gear train in words (even for an engineer).  However, if you study the link you will better understand gears and make your own calculations which is ideal for a school project, right?
635462
Maybe the car designers like a bottom heavy babe! In fact CG or centre of gravity of the vehicles is at a low point if the heaviest weight is aaranged at the lowest point. Since engine is the heaviest part, it is placed at the lowest possible point to give car highest stability
37821
If you do that a lot, the clutch release bearing will wear much faster. You'll know when its wearing as there will be a buzz or hum when you press the pedal down which wasn't there before.\n\n:-)\nDC
1036638
http://www.sheldonbrown.com/gears.html
559385
Too many variables to give a complete answer. It depends on how steep the hill is, how fast you are going, how much you car weighs, how much torque your engine develops, etc\n\nThe short answer is: If your engine is lugging down, switch to a lower gear. If your car has an automatic transmission, it should select the appropriate gear automatically.