A 100 kg car is on a 30m radius banked frictionless corner moving at 10m/s--physic----->Please help!!!!!?
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)
1044764
The law of conservation of momentum holds in this case.\n\nThe initial momentum of the system must equal the final momentum of the system.\nIn this case, we will call the "System" both cars.\n\nCar 1 has a mass (m1) of 1000 kg, velocity (v1) of 20 m/s.\nCar 2 has a mass (m2) of 800 kg and a velocity, in what we will call the negative direction, v2 = -30 m/s.\n\nThe total momentum (P) does not change.\n\nP = P1 + P2\n\nP1 = m1 * v1 = 1000 kg * 20 m/s\nP1 = 20000 kg m/s\n\nP2 = m2 * v2 = 800 kg * -30 m/s\nP2 = -24000 kg m/s\n\nP = 20000 kg m/s + -24000 kg m/s\nP = -4000 kg m/s\n\nSo the resulting 1800 kg mass will have a momentum of -4000 kg m/s, meaning that it will be traveling in the "negative" direction, AKA, South.\n\nP = -4000 kg m/s = 1800 kg * v_f\nv_f = -4000 / 1800\nv_f = -2.22 m/s\n\nThe resulting blob of the two cars which stuck together will be traveling -2.22 m/s South.
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
919282
please make your question more understandable\n\nif it decreases from 30mph to 15mph over 75 metres then at this deceleration its logical to say it would take another 75metres to stop\nbut something tells me thats not the answer ! if you want me to do your homework at least have the decency to type in the full question.
784770
Consider the forces and directions of the forces on the car:\n\nThe force of gravity pulling the car directly downward\nThe centrifugal force pulling the car directly horizontal\nThe force of the car on the road, which is banked\nThe friction of the road on the tires.\n\nSince it is icy, let's set the friction to zero\n\nThen the forces parallel to the road are the parallel component of the centrifugal force:\n\nm*cos(3.4)*r*w^2\n\nand the force of gravity towards the center\nm*g*sin(3.4)\n\nWhen these force balance, there is no force of friction parallel to the road surface.  So \n\nm*cos(3.4)*r*w^2=m*g*sin(3.4)\nw=sqrt(g*sin(3.4)/(cos(3.4)*r))\nThe linear velocity, or what the driver will read on the speedometer\nis w*r\n\n8 m/s\n\nj
756380
Firstly, let me point out to you which answers are wrong and which answers are right. Otherwise you'll be just as confused as you were to begin with.\n\nThe first answer is confused. The second and third answers have no idea of what their talking about. The other answers are correct.\n\nHere's why the first answer is confused:\n\nSpeed and velocity use exactly the same units. \nSo it is perfectly acceptable to say "a car travels around a sharp curve a 60 km/hr". \n\nRestated, what we are saying is that the car moves around a curve at constant speed of 60 km/hr. The statement above does not include directional information. \n\n(However, suppose we said "a car travels around a sharp curve with constant velocity of 60 km/hr". Then, clearly this would be wrong. Velocity does change as it goes around the curve.)\n\nSecondly, let us recall some basic facts of 'Uniform Circular Motion' \nAn object that travels with Uniform Circular Motion has:\n- constant speed \n- changing velocity  \n- acceleration towards the centre\n\nNow, the car in question displays uniform circular motion. This is so even though we are considering a partial circular path.\n\nSo, we can use this import formula (you should memorise it) \n\n    ar = v ^2 /r\n\nThis relates the centripetal acceleration, ar, to the radius of the circle (or in the case of a partial circle, the radius of curvature), r.\n\nFrom the formula you can see that a sharp bend, i.e. a small r, will give a larger value of centripetal acceleration compared to a gentle bend, i.e a large r. \n\nSubstitute v=60 and different values of r into the formula above to verify this.
725445
You can't without the radius of the wheel...m/s is a tangential speed.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
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.
782512
You need to use kinematics and Newton's 2nd law to solve the problem.  In the horizontal direction, Newton's 2nd law states:\n\nForce - friction = m*a.  Since you are given the values, you can solve for the accleration:\n\n1149 - 909 = (2700)*a, solving for a, you get:\n\na = 0.089 m/s^2\n\n----------------\n\nNow use kinematics to solve for the distance needed to reach the desired final velocity.  Use the following kinematic equation:\n\n2ax = v_f^2 - v_i^2.  You know that:\n\nv_f = 3.2 m/s\nv_i = 0 m/s (starting from rest)\na = 0.089 m/s^2\n\nNiw solve for x:\n\nx = [v_f^2 - v_i^2]/(2a) = 3.2^2/(2*0.089) = 57.6 meters\n\n---------------\n\nHope this helps
841492
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
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
974269
To stay on the road with no slip the vector of total force on a car F must be strictly normal to the bank with b=2.5 degrees. Vector F=Fx+Fy, Fy=m*g being the weight of the car, Fx=m*a=m*r*w^2 is centripetal force on the curve of r=120m, w=v/r is angular speed, v is linear speed, a=r*w^2 is centripetal acceleration. On the other hand as Fx=F*sin(b) & Fy=F*cos(b) or Fx/Fy=tg(b) or [m*r*(v/r)^2]/[m*g]=tg(b), hence v=sqrt(g*tg(b)*r)=7.17m/s=25.81 km/h. Hi to SS!
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)
771460
Since the crate is moving at constant speed, acceleration = 0, net force is 0.\n\nApplied Force 150 N --> [41.5 kg mass] <-- Friction F = 150 N\n\nFor sliding friction, F(friction) = u N , where u (mu) is the coefficient, and N is the Normal force (imagine an arrow pointing down from the mass).\nN = mass x g
970605
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
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!