A car travels at a constant speed around a circular track whose radius is 2.1 km.?
98015
The answer to the second question is .02s for one revolution.
900310
A wheel mounted on a wall is rotating. Every second, its velocity advances 4 radians/second, which is equivalent to 229.183 degrees/second. That is, if it is moving at 0 degrees/second at time 0, it will be moving at 229.183 degrees/second after 1 second, and it will be moving at 458.366 degrees/second after 2 seconds. This is actually the answer to part (a).\n\n(a) The angular speed at t=2 seconds is 458.366 degrees/second, or 8 radians/second.\n\nWe assume that the wheel is rotating counter-clockwise. That is,  if you look at it, a point at the top of the wheel will move to your left.\n\nAt t=0, we focus our eyes on a point on the rim of the wheel. That point is located 57.3 degrees above the horizontal. That is, if the wheel was a clock face, the point would be located around 1 o'clock. 1 o'clock is located at 60 degrees above the horizontal, so the point is actually located at where the hour hand would point at 1:05, which is just passed 1 o'clock. (keep in mind that 3 o'clock is located at 0 degrees and 12 o'clock is located at 90 degrees)\n\nNow, at t=2 seconds, we already found that the wheel is rotating at 8 radians per second. Since the radius of the wheel is 2 meters, if the wheel moves 2*pi radians, the wheel moves 2*pi*r meters. Thus, there are r=2 meters per every radian. Thus,\nthe point at the edge of the wheel is traveling around the wheel at (8 radians per second)*(2 meters per radian) = 16 meters per second. Similarly, since the wheel is accelerating at 4.0 radians/second/second, the point at the edge of the wheel is accelerating along the circle that is the edge of the wheel at at (4.0 rad/s/s)*(2 meters/rad) = 8.0 meters/s/s. \n\nHowever, there is also an acceleration normal to the edge of the circle pointing toward the center of the circle. That acceleration is thus a "centripetal" acceleration, and is given by v*v/r, where "v" is the tangential speed of the point. We know the tangential speed is 16 meters/second, so the centripetal acceleration is 128 m/s/s. \n\nThese two accelerations (tangential and normal/centripetal) are perpendicular to each other, so the TOTAL acceleration is the length of the hypotenuse of the right-triangle formed with each of them on a leg. That is, the total acceleration is the square root of ( (128 m/s/s)^2 + (8 m/s/s)^2 ), which is 128.25 m/s/s.\n\nThus, the answer to (b) is:\n\n(b) The tangential speed is 16 meters/second. The tangential acceleration is 8 meters/s/s. The centripetal acceleration is 128 m/s/s. The TOTAL acceleration is 128.25 m/s/s.\n\nFinally, you want to know the angle where P is after 2 seconds. Since everything started at rest, you know that (you can derive these with a little calculus; keep in mind that acceleration is change in velocity over time and velocity is change in position over time):\n\nacceleration = 4 rad/s\n\nvelocity = acceleration*t = (4 rad/s)*t = 8 rad/s\n\nDISPLACEMENT = 0.5*acceleration*t^2 = 0.5*(4 rad/s)*t^2 = 8 radians\n\nSo the angle of point P has moved 8 radians, which is equal to 458.366 degrees. Since the point P has started at 57.3 degrees, then you just add 57.3 degrees to this number and you get 515.666 degrees. Now, there are only 360 degrees in a circle, so this means that the point has completed one revolution and has gone 155.666 degrees into the next circle. Thus, your answer to (c) is:\n\n(c) The angular position of the point P is 155.666 degrees.\n\nI hope that helps.
336809
On a flat curve, the centripetal force on the car is merely the friction between the tires and the road. Note, however, that this friction is STATIC friction, which has a maximum value. \n\nF(radial)=f(static)=mv^2/r\n\nIf the velocity exceeds a certain value, the car will slide. \n\nOn a banked road, however, one component of the normal force will point inwards, adding to the centripetal force. This allows the car to take the curve at higher velocities without sliding:\n\nF(radial)=f(static)+normal*sin(angle)=mv^2/r
763022
Use very narrow and small radius wheels on your car. the rotational inertia must be kept to absolute minimums so that most of the energy from the ramp will go into velocity of the car rather than into the rotation inertia of the wheels. this will provide the greatest speed down the track and assuming the track is short the you will get there quicker. If however the track is long you will want to do the opposite by making the wheels large to store the energy and when the car hits the flat section it will not be going as fast but it will go a lot further\n\nE-mail me and i will be happy to explain various other little details  but the main emphasis must be placed on the wheels and knowing the track length is very important to the design and outcome.\n\nI will be able to tell you actual velocities expected at the bottom of the ramp and can calculate the expected distance the car will travel if i am given a little more information. in particular the size of the wheels, mass of the wheels, mass of the car and just so you know the aerodynamics are not going to be any issue at all because the relative velocities are well below speeds that need to consider design shape or aerodynamics. usually speeds in excess of 40mph are where dynamics become an issue and need to be considered. \n\nWell travis is incorrect in the friction issue.  You obviously want the wheels to spin freely but any extra emphasis on this will not gain any performance.The fact that all of the cars are going to be essentially the same with the coefficient of friction then the rubbing area makes absolutely no difference at all. This is because the friction is related to mg*(coefficient of friction) and area has no effect. I will also be happy to elaborate on this point a little more and explain a lot of fallacies friction,tires,roads, brakes and why contact area has essentially no effect.
217138
the particles move faster.
262728
Simple.\nThe car tyres rotate thus the force is called centripetel force(f).It is equal to v^2/r where r is the radius of tyre.\n\nforce at 15m/s = 15*15/r\nforce at 30m/s = 30*30/r\n\nWhen we divide them we get 1:4 the answer\nit is assumed that car tyre is of same radius in both cases.
598306
nothing-\nthe system of atoms inside your car is stationary relative to the baloon.
725445
You can't without the radius of the wheel...m/s is a tangential speed.
741402
a) F = ma \nYou're given F = -6070 N (negative because it's in the opposite direction) and m = 2600 so a = -2.33 ms^-2. Then over 1.76s, the change in velocity is (-2.33)(1.76) = -4.10 m/s. Car velocity is then 34.9 - 4.10 = 30.8 m/s.\n\nb) v^2 = u^2 + 2as\nv = final velocity = 30.8 m/s (from above)\nu = initial velocity = 34.9 m/s\na = acceleration = -2.33 ms^2 (from above)\nRearrange to get s = (v^2 - u^2)/2a = 155.61 m\n\nc) v = u + at\nYou want v = 0. So if the force is constant then acceleration is constant at 2.33ms^-2. \nu = initial velocity = 34.9 m/s\nSo rearrange to solve for time: (v-u)/a = t = 14.98 seconds.\n\nHope that helps. And learn your equations!
771612
hello\n\na)take the momentum of the sports car and divide by the total mass of  both cars. 25×920÷3220=7.14m/s\n\nb)using the same method as a), we can find the answer. but use a little logic here, if the total mass doubles then the final speed must be halved for conservation of momentum to be followed. v=12.5m/s\n\nc)KE=0.5mv². speed is halved. v² means kinetic energy should be quartered (0.5²). but mass was doubled due to there being 2 cars now. so overall kinetic energy has only been halved (0.5²x2). the other half is lost.
864537
r1 * A1 = r2 * A2\n0.21*(pi*1.2^2) = r2 * (pi*0.23^2)\nr2 = 5.72 m/s
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
483925
You may want to try Centripedal Force i.e. Turnings and slopes..\n\nBut it has some stuff to do with Physics as well. \n\nAlso you may want to find the maximum and minumum speeds at turnings. Include Frictional Force and try to minimize it using Calculus. \n\nTake Simple Circuits like circles, oval, parabola etc. Then take their equations and use calculus to find area under the curve etc. etc.\n\nWell thatz all comes to my mind for now!! :) BEST OF LUCK..
1170380
You have mass of car = 1500 kg.\nand speed = V = 25 meters per second.\n\nYou will recall the equation \nE = (1/2) * m * v^2\n\nKinetic Energy = 0.5 * mass * speed *speed.\n\nPunch these numbers into your calculator and you will get a numerical answer with units kilogram meters meters per second per second.\n\n\nThis is the kinetic energy of the car, which is removed by the brakes.\n\nI'm not going to do all your homework for you.  Just the first part to get you rolling.\nHave a great day.
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)
494819
MORE OR LESS 135 KM/H.