Is mileage in your car figured by wheels turning or by something in the engine revolving?
439803
Yeah, attached to wheels and calibrated according to size.\n\nWorks fine unless you are a skidder on the road, and cover a sizable portion of your journeys with the wheel locked. ;)
325034
If you are using an absolute reference to f4, then all you have to do is match back up the correct mileage you want to multiply the mileage rate by. If they are listed vertically, with f9 being the first cell, you can just click and drag down, the forumla should look at f10, f11, f12, ect while still maintaining the same rate in f4. \n\nIf not, you are going to have to edit the formula after you copy to reference the correct mileage.
18946
Reset your odometer when you completely fill your gas tank. Then the next time you fill up, check the number of miles you drove and also the gallons of gas you put in your tank. \n\nDivide the miles you drove with the number of gallons you put in your car and you got the mpg for your car.
496203
without friction, the car will not move when the wheel is turning. without friction, the wheel will not stop when u step on the brake. even if it stops, the car will not stop. without friction, u can't hold anything in your hand. u cannot stand upright. your glasses will not stay on your face. u will slide off your chair. everything would slide around.\n\nwithout friction, less energy is loss as heat and noise. less energy would be needed to power an engine. tire would not wear out. thing in motion would move longer.\n\nu get the idea
1046019
when a car speeds by you, you feel wind. it's the same way with the earth, as it spins very fast.
1184449
It does not matter how many seconds your car moves in first gear, after that it should be 60 miles/hour and the feets you do the math by yourself. GOT IT???DUHHH
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.
392981
Is the speed at which it travels.
848599
No. The velocity of the wheel is changing as the car moves (because velocity indicates direction). So as the car moves forward, the velocity of the surface of the wheel is continuously changing.
232758
Normally when car manufacturers are talking about just HP, they are talking about the horsepower the engine makes at the flywheel. Wheel HP on the other hand is the measurement of horsepower at the wheels. Between the flywheel and the drive wheels, horsepower is lost in components such as the transmission and the differential. So the HP an engine makes at the flywheel is always going to be higher than the horsepower reading at the wheels.\nAnd when it all comes down to it, the horsepower made at the wheels is what really moves the car.
784055
The way to do this is to first work out the acceleration required to turn the car round the curve.\nFrom there it is a simple matter of working out what angle the road needs to be at so that gravity will provide that acceleration.
257674
RMPH stands for Rental Market Provincial Highlights which is a magazine.\n\nRPM x 1000 which is displayed on the gauge face of your car means Revolutions Per Minute which is measured by how many times your flywheel or crankshaft can revolve around its axis in one minute.  Since an engine spins very fast (most engines are about 800 RPM at idle) engineers have decided divide the numbers up instead of spelling them all out (ie: 1000, 2000, 3000).\n\nAs your engine spins faster your RPM will increase along with your fuel consumption and emissions production.  Saying that, it would make sense that if you have a standard or manual transmission in your car, you would be using more gas in third gear at 55 MPH compared to fifth or six gear at 55 MPH.  The change in your gas mileage, which is a concern these days, is little to none over a course of a week if your engine RPM is at 2,500 or 3,000.  However, if you calculate your total gas consumption based on two different RPM tests (driving one week with high RPMs and one week with low RPMs while recording gas mileage) over a course of a year you will see a dramatic consumption based upon the fact that if its accumulating, its well... accumulating.\n\nAnd no, the numbers do not mean slower pick up/acceleration.  The numbers are for viewing.\n\nHope that helps!
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
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
18418
What I always do when I'm about to go on a road trip is to get driving directions from the starting point to the destination (recommend Google maps or Yahoo maps). Below the "From" and "To" addresses it normally lists the mileage. For example, a trip from Milwaukee to Charlotte is 901 miles. Then you divide that number by your cars average MPG. I suggest overestimating a little (e.g. if my car's average mileage is 29 miles/gallon, I'd use 26 or so to be safe). Then you multiply that number by the average price/gallon you expect to pay (lets say about $2.50). To make my examples even more explicit:\n\n[total distance in miles (901 mi)] ÷ [car's MPG (26 mi/gal)] x [gas price per gallon ($2.50)]= [Total cost of gas ($86.64)]. \n\nHave a great trip!
176123
The force that pulls you towards the side of your car when you go around a corner too fast is actually inertia.  It's the same force that makes you fly forward when you hit the brakes too fast or sink into your seat when you lay on the gas.  According to Newton (I believe it's his third law), "an object in motion tends to stay in motion and an object at rest tends to stay at rest."  Well when you are in your car and you are moving in a straight line, your tendency is to keep moving in that straight line.  When you take the turn too sharply, the car is turning but your body's natural tendency is to keep moving in that straight line, so you get pushed up against the side of the car.  The same justification explains why you move forward or back when the car accelerates or decelerates too quickly.\n\nCoriolis force is a completely different force that is caused by the rotation of a sphere and the fact that the sphere is densest in the middle, with decreasing density as you approach the poles.  This change in density causes a force which makes moving things tend to move towards the equator of the sphere.
51533
You are driving a car.  The car in front of you stops\nsuddenly.  You hit the brakes.  Tires squeal, but the car still moves. Inertia.