what is the basic design of the hydraulic car jack? what feature does the jack have to allow it to hold the car without being lowered? also, what are the different calculations that must be done on this jack?eg. buckling etc.
128716
If you have a small hydraulic piston pump where the cylinder has an area of say one square inch and you can drive its piston with a force of say 100 Lbs (using a lever arm), the pump will produce a hydraulic pressure of 100 PSI (pounds per square inch).  If the hydraulic fluid is delivered to a larger cylinder and piston (hydraulic ram) it will produce a force equal to the area of the larger piston time the supplied pressure.  If the larger piston has an area of say 10 square inches, the force will be 10 (square  inches) x 100 (pounds per square inch) = 1,000 pounds.  If you pushed down on the jack handle with 10 pounds of force, using its leverage, your 10 pounds of force has been magnified to 1,000 pounds of force (enough to lift one car tire off the road?)  A tiny check valve allows you to jack up the car with successive pump strokes by preventing backflow of tthe hydraulic fluid.  When you want to let the car down you (twist the handle?) open a bypass valve around the check valve.
198940
Your all wrong, it's the gear shifter.\n\nThese cars allways do this, if your on a hill it's worse.\n\nHave a person-persons push on the car to "unstress" the pressure on the trans, then move the shifter into park.\n\nThis will work.
675195
The basic principle is to use leverage. Here's one way off the top of my head (I wish I could draw a diagram):\n(1) Hang the object from a rope/cable that runs up through a pulley. \n(2) Fix the other end of the rope so that it pulls up on a lever arm. \n(3) Make the other end of the lever arm (beyond the lever's pivot point) push down on the scale. \n(4) Since you want to effectively reduce the weight pushing on the scale, the length LR of the lever arm between the pivot and the rope will be shorter than the end LS between the pivot and the scale. \n(5) The relationship between the weight WO of the object and the weight WS you read on the scale will be: WO = (LS/LR)WS. \n(6) This ignores the effects of mechanical imperfections and friction in this apparatus. \n\nAlso, I can imagine several other mechanical configurations that would achieve the same thing.
130796
Work = force * distance\n= 1000 N * 10 m\n= 10000 Nm\n\nYou wouldn't take into consideration the mass of the car because the force you are applying is horizontal (assuming that you are pushing straight on the car and not down at an angle).  The only time you would use the 3000 N is if you took friction into consideration.  In that case, you would have a friction force opposing your 1000 N, thus reducing it.  (Force of friction = coefficient of friction * normal force)\n\nPower = work / time\n= 10000 Nm / 100 s\n= 100 Nm/s
278466
First Question:\n\na) Solving for the height of the 1st brick\n\nh=1.00m (sin 30.0)=0.50 m\n\nSolving for the velocity of the 1st brick at the bottom\n\nKE at the bottom = PE at the top\n1/2 v^2 = (weight) h\n1/2 v^2 = 24.3 (0.50)\nv = 4.93 m/s\n\nb) momentum before = momentum after; momentum is just m*v\n\n(24.3/9.8)(4.93)+(35.7/9.8)(0)\n=[(24.3/9.8)+(35.7/9.8)] [v]\n\nv = 2.0 m/s\n\nc) F = ma\n-5.0 = [(24.3/9.8)+(35.7/9.8)] a; F is opposite to motion\na = -0.817 m/s^2; deceleration\n\na =(vf-vi)/t; vf=final velocity; vi=initial velocity\n-0.817 = (0-2.0)/t\nt = 2.45 seconds\n\nd) 2ad = vf^2 - vi^2\n2(-0.817)s = 0^2 - 2^2\nd = 2.45 meters\n\nSecond Question\n\na)Make West East axis as the x-axis; N S axis as y-axis.\n\nx momentum before = x momentum after\n\n880(0)+1585(12) = (880+1585)(vx)\n\nvx = 7.72 m/s; x component of final velocity\n\ny momentum before = y momentum after\n\n880(-16)+1585(0) = (880+1585)(vy); compact car is headed in the negative y-axis.\n\nvy = -5.71 m/s\n\nv^2 = vx^2 + vy^2\n\nv^2 = 7.72^2 + (-5.71)^2\n\nv = 9.6 m/s answer\n\nangle = Arctan 5.71/7.72\n\nangle = 36.5 degress south of east\n\nb) F = coef of friction (weight of the cars)\n\nF = 0.58 (880+1585)(9.8)\n\nF = 14011 N\n\nF = ma\n-14011 = (880+1585) a; - because opposite motion\n\na = -5.68 m/s^2; decelerating (-)\n\n2ad = vf^2 - vi^2\n\n2(-5.68)d = 0^2 - 9.6^2\n\nd = 8.11 meter from the poin of impact
420645
OK, I'll try to help you learn the principle so you can pass the test. My explanations are in ().\n\nA high tech jet car speeds along the desert floor at a constant velocity of 180 miles/hr north.a) what is the net force on the jet car? the jet car hits some sand which causes a drag force on the car of 5000 N to the south. The computer instantly compensates so the jet exerts a thrust force on the car of exactly 5000 N to the north at the same time the car hits the sand. The jet car now... choose ALL that apply: \n\n(There has been drag all along. Since the car has been going at a constant velocity, the car's jet has to be exerting an equal and opposite force to compensate for the drag. If the drag and thrust are equal and opposite there's no net force. Newton's 1st says it will stay in motion at the same speed in the absence of a net force. So now, the net force = 5000 N - 5000 N = 0.) The car \n\nmaintains constant velocity\nmoves with constant non-zero velocity\n\n(I explained all this above.)\nNewton's first law states: When the __ net force__ on an object is zero, the object will __maintain__ constant __velocity__.\nAnother way of saying this is that, if the total force on it equals zero, an object's _acceleration__ will also be zero.\n\nIs it necessary to apply a force to keep an object going in a circle at constant speed? \n\n(The object will want to keep going at constant speed in a straight line because of inertia - without a net force it will do that. The force required to go in a circle is towards the center.)\n\nYes, because the __velocity__ is changing.
661052
The one thing that made the elevator or lift a possibility for comercial use was the invention by Mr. Otis of the mechanical break system that prevents the car from falling even when the cable is cut.  It basically drives a wedge against the wheels and the rails when the car starts to go down faster than it should. \n\nPower outages pose no threat since the car is counter balanced by a heavy counter weight and the motor and gears simply stop in position with no car movement during a power outage.\n\nThe biggest concern is, what happens if a cable snaps?  Mr. Otis toured the country demostrating his brake system by acutually getting in a car, having it raised and then he personally chopped through the rope suspending it.  Some cars will lurch and drop a couple of feet then stop and lurch again and will keep doing this until the car is finally stopped at the bottom of the shaft.  This kind of drop is really scary but no one has died from a falling elevator due to a cable failure as far as I can remember.
220506
the head phone jack is supposed to shut off the spaeakers when you plug in the head phones. the contact in the phone jack might be bent out of shape. this happens when crazy kids wiggle the phone plug side-to-side and ruin it. a technician can bend it back in place but i imagine they would just replace it to save time. until you get it fixed just turn down the speakers.
305474
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
84456
Car manufacturers also own the towing services that come and take us to the nearest gas stations that OH, incidentally, the car manufacturers own too. So with this monopoly going on they gouge us with the gauge since we over estimate how long we can go especially on that last quarter of a tank of gas. So in short......it's a racket. A gauge gouging racket. :D
299159
A Button - activates the four hydraulic jacks on the undercarriage (built so the car could be lifted for maintenance, but also used to give the car short-distance leaping ability).\n\nB Button - Activates special "super-grip" tires that allow the Mach 5 to cling to most surfaces.  Also activates auxiliary engines to generate extra horsepower to those wheels.\n\nC Button - Activates the front mounted saw-blades to cut through obstacles.\n\nD Button - Seals the cockpit with a bulletproof and air-tight dome (air is then pumped in by a sealed oxygen canister system).\n\nE Button - Activates special headlights that work in tandem with the special goggles on Speed's helmet to give him infrared vision.\n\nF Button - Converts the Mach 5 to minisub mode.\n\nG Button - Launches the homing robot found in the car's hood.
538345
Use the fact that PV/T = a constant, we find the constant by pluging in the values we have, T = 20C, but we must convert that to K first so 20C = (273 + 20) = 293K.  And 1 atm should be converted to physicall units of Nt/m^2 for an easier solution.  1 Atm = 101325 Nt/m^2.    So our constant is (101325)(4.5)/293 = 1.336 x 10^7 so now you have one equation as\n\nPV/T = 1.336 x 10^7 setting T = 260C or 533K gives\n1)  PV = 7.121 x 10^10\n\nThe next equation that we need is the force on the cylinder end given the pressure, which would be\n\n2)  .01P = F\n\nThe spring will stretch a distance 3)  d = F / (2.00 x 10^3)  \n\nThe volume changes by  4)  V = 4.5 + d * .010   \n\nThis gives you 4 equations which you can now solve to find the solution for V, d, P and F
165707
It could actually help another vechicle, I don't know about a large truck(semi?) but here read this\n\nThis slipstream effect works in the following way. As a single car races around a track, it creates a "bubble" of high-density air in front of it and a bubble of low-density air behind. The difference in the pressure between these two air pockets creates drag, the force that impedes motion. This drag force limits the top speed the car can attain. But if a second car pulls up behind the first, the slipstreams created by the two merge so that the first car losses its aft bubble and the second car loses its front bubble. This effect reduces the drag force each car experiences and both are able to travel slightly faster. \n\nhttp://www.aerospaceweb.org/question/aerodynamics/q0092.shtml
53519
YES!  It's perfectly possible to _build_ such a car.  If the magnet is just attracting the car, it would pull forward on the car, and the car would pull back on the magnet.  Nothing moves.  If the magnet is attracted to an object ahead, the rig would move forward but would stop when the magnet got there.
126123
You can bend the flow of water from the hose by swinging it on the air.\nIn fact you cannot bend its properties called H2O. \nYou cannot bend it either using your hand or vice grip.\n\nWater is not compressible either.\n\nHydrokinesis on only a fiction it could not be true.
552327
get a trolly jack, jack up the front of your car and push it into a busy road. the police will make them remove the clamp. they cant prove that the car was not there when it was clamped.\nthis works. i have done it myself.\ngood luck.
79957
Sticking by the rules you specified;\n\nConstruct an inclined channel (ramp) where the 1 kg (2 lbs)weight (hopefully a metal ball) falls and accelerated by gravity. place the car that will catch and hold the ball at the end of the ramp. The falling ball's inertia will propel the car almost equal to its speed and may reach more than 1 meter on a flat even surface, design the car with low center of gravity, with less than half the weight of the propellant ball (weight) as a consideration.\n\n************* \n\nAh, New data acquired --- resolution: same application of inertial force using a rigid pendulum.\n\nConstruct a pendulum with the required height putting the fulcrum point at 10 cm directly above the floor of the car, giving you a swing radius of 10 cm from top to bottom, use a cardboard/plastic tube (shute) to carry the pill bottle but loose enough to set it free after a sudden stop once it reaches the bottom where the position of the pill bottle is parallel with the floor of the car. This configuration may not give you as much distance relying on gravity alone but may be able to get 1 meter if conditions are favorable, pertaining to weight of car and means of motion (wheel and shaft should be well oiled for free-wheeling), relevant to weight of falling object, unless you are allowed to spring load the fulcrum. You can put the fulcrum as part of the car (but it will add more weight/ but you wont need the shute, all components will be integrated), or as a separate launcher (which you will need the shute). make sure the center of gravity on the car is low for stability. when integration of all component is involved make the wheel base equally wide as 10 cm. If just for the pill bottle, just narrow enough to carry the bottle.\n\nmy idea: http://server4.pictiger.com/img/124010/other/1-kg-inertia-car.jpg\n\nInnovation is part application and part logical creativity. Isn't science great!\n~meow