A coffee cup on the dashboard of a car slides forward on the dash when the driver decelerates from 41 km/h to?
954566
You have to know the coefficient of friction between the cup and the dashboard and the rate of deceleration. When the cup's forward inertia is more than the friction holding the cup there, it slides.
1020689
The figures are (for the 2005 Renault R25):\n\n0 to 100 km/h: 1.9 seconds \n0 to 200 km/h: 3.9 seconds \n0 to 300 km/h: 8.4 seconds, may be slightly more or less depending on the aerodynamic setup. \nThe acceleration figure is usually 1.4 g (14 m/s²) up to 200 km/h, which means the driver is pushed back in the seat with 1.4 times his bodyweight.
768384
because a car turns heat energy into mechanical energy
984567
a=(delta V) /t = (26/24)=1.08 m/s2\n\nF= mxa,   \n\nwe know weight of car is 11420 N, we need mass\n\nm=11420 / g= 11420 / 9.81 = 1164 kg\n\nso F= 1164 x 1.08 = 1257.25 N to stop the car\n\nfriction = F/Fn =  1257.25/11420 = .11
54242
A change in a objects velocity produces acceleration.\n\nAcceration is a measurement of the rate of change of velocity, and is usually measured in "meters per second squared" or ms^-2.\n\nIf a car is travelling at 5 meters per second, and over a time of 5 seconds, increases its speed to 15 meters per second, it has accelerated at a rate of 10 meters per second over 5 seconds, or 2 ms^-2 (2 meters per second squared)\n\nHope that helps
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
789497
Between 30 and 40 mph.
91916
a car trvelling in a straight line\na beam of light
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
149924
it will hit the back of  the car
12602
If I understand the question correctly, I think the force you're looking for is inertia. Your body is moving at the speed of the car - then the car stops. Now the car has no speed, but you're still moving forward at that speed until something (hopefully the seat belt) acts to stop you.\n\nIf you're referring to the force provided to stop you - that's the Newtonian counter-force to your inertia moving forward.
114124
Probably the simpliest way I can think of is just to strap the balloon to a set of wheels and position it so that the opening is point in the opposite direction you want the car to move.  When the balloon is released, the air can escape out the hole and propell the car forward.
664508
because of inertia it will be traveling at 60mph as it leaves the car\nbecause of the fly's body shape theoreticaly it would give it lift assuming that the fly was facing foward\n\nrealisticly it was facing sideways, so the air would be traveling over its body sideways it would have been plummeted into the ground and its wings would have been ripped off
135305
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.
884398
The work done on the car by the net force equals the change on the car's kinetic energy.\n\nWork = Force * distance\nKE = 1/2 mv^2\n\nWe are given a mass for the car and the final velocity, so from this we can calculate the car's final KE,\nKE = 1/2 (2000 kg) * (2 m/s)^2\nKE = 4000 Joules\n\nSince in this case, KE = W\nW = 4000 Joules = F * d\n\nWe know the net force to be the sum of all the forces acting on th car.  Since there is a "forward" force of 1140 Newtons and a "backward" (what we will call 'negative') force acting on the car of 950 N, the net force is 1140 - 950 = 190 Newtons\n\nW = 4000 J = (190 N) * d\nd = 4000 J / 190 N\nd = 21.05 meters
782291
Assuming that the friction with the ground provides a constant force, the clumped-together vehicles decelerate at constant rate until they stop. Because we assume the ground to be horizontal, that rate is (mu) * g, where (mu) is the coefficient of friction. Therefore the rate of deceleration is .923 g or 9.052 m/s^2.\n\nThe distance traveled is 18 m. For constant acceleration (or deceleration, it does not matter as long as the velocity at one end of the time period is zero) x = 1/2 a t^2. We know x and a; therefore we find t = 1.994 sec.\n\nNow we use this time and the assumption of constant acceleration to find that the initial velocity was v = a * t = 18.05 m/s. Therefore the intial momentum was 18.06 * 3000 = 54 150 kg m/s, directed toward the northwest.\n\nTo cause that momentum, the momenta of the individual vehicles added, and they are vectors. The truck was headed north, the car, west, and they ended up going northwest. Therefore they had equal-magnitude momenta. (The tangent of the resultant angle is the ratio of the two.)\n\nSay Pt = truck momentum = Vt Mt, Pc = Vc Mc.\nDealing just in magnitudes for the moment, \nPt = Pc --> Vt Mt = Vc Mc, so Vc = Vt (Mt / Mc) = 2.75 Vc.\nBecause the vehicles started at right angles to each other, the magnitude of the final momentum is Pf = sqrt(Pt^2 + Pc^2) = \nsqrt((Mt * Vt)^2 + (Mc * 2.75 * Vt)^2) = 3111 * Vt\nWe have already found Vf = 18.05, and we know the combined mass is 3000 kg; therefore we infer Vt = 18.05 * 3000 / 3111 = 17.41 m/s. Therefore Vc = 47.87 m/s.
861510
The way to solve this problem is to decompose the forces acting on the car, but not in the x and y direction. Instead, decompose the forces in the direction of the incline and the normal force.\n\nHence, we have:\n\nF(friction) = μN, moving up the incline, and μ is the coefficient of static friction\n\nThen we have gravity acting on the car, and the component that's acting down the incline is = mgsinθ, where θ is the incline of 15°.\n\nFor the car to be safe in making this turn, the forces along the incline must be equal:\n\n(a) F(friction) = μN = mgsinθ = (1450)(9.8)sin(15°) = 3677.82 N\n\nBut the forces acting perpendicular to the incline on the car must also be zero:\n\nN = mgcosθ\n\nNow substituting this into the previous equation, we get:\n\nμN = μ(mgcosθ) = mgsinθ, this implies that the coefficient of static friction is:\n\n(b) μ = sinθ/cosθ = tanθ = tan(15°) = 0.268