what netforce is needed to maintain the constant velocity of a car moving in a straight line? explain?
144016
The net force needed is zero.  Force is mass * acceleration, and acceleration suggests a change in velocity.  For constant speed and direction, the acceleration is zero, and hence the net force is zero.  When moving at constant velocity, the frictional and drag forces (which work to slow the car) and the drive force of the engine are exactly equal and opposite.  The net force is the sum of these two opposite forces, and hence the net force is Zero.
198734
The force of friction is dependent on two things, the Normal force between the "object" and the "surface" and the coefficient of friction between the two.\n\nThe normal force on a level surface is equal the weight of the object.  You are given the mass, from this, you can easily calculate the force of weight (and this F_n).\n\nYou are given the coefficient of friction, greek letter mu.\n\nThe force of friction can be found as,\nF_f = F_n * mu\n\nThe force of friction is always in the direction opposite that of the motion.  So if your traveling "forward", the frictional force is pointed "backward".\nWhat are the two components in the formula that affect the magnitude of the frictional force?  Just the normal force (which is dependent on mass and gravity) and the coefficient of friction.  For all intensive purposes, mu is constant, as is the mass and gravity of the object.  Therefore, the force of friction is constant.\n\nYou are given in initial velocity.  You are trying to find how far (change in distance) the car will travel while this constant force is being exerted on it before the car comes to a stop.\nThere are several ways this could be found.  I think the energy method would be the easiest, but I will cover the kinematic method as well....\nThe car has some initial kinetic energy when the problem starts out.  KE = 1/2 mv^2.  After the car comes to a stop, the car has zero KE...therefore some work had to be done on the system to change the energy (we assume a level road).  The change in KE equals the work done.  Work is also equal to the product of the force applied over the distancd applied.\nW = F * d\nSince the frictional force is what is causing this change in energy (which we know, it was calculated earlier), and we also know the change in KE, we can simply solve for the distance, d.\n\nThere is also another possible solution.\nWe know that the frictional force will cause an acceleration.\nF = m*a\nWe know that an acceleration will cause a change in velocity.  We know ,\nv_f^2 = v_i^2 + 2ad\nwhere v is velocity (i is initial, f is final), a is acceration, and d is distance.  Simple solve for d.
684454
Let's consider the case when the Car is right in front.\n\nAt that instance, the car is travelling tangentially to a circle of radius 132ft, where the Camera is the centre of the circle.\n\nCircumference of the Circle = 2*(pi)*R\n= 2 * pi * 132 = 264*(pi) ft.\n\nTime taken to travel the entire circumference by the car\nT = circumference/velocity = 264*(pi)/264 = pi seconds\n\nHence the car would require pi s to cover 360 degrees of the circle.360 degrees = 2(pi) radians.\n\nW = angular velocity = 2(pi)/pi = 2 radians/sec.\n\ncan't help about half a sec later now
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.
553994
1st - An object in motion/rest will stay in/at motion/rest unless acted upon by another force.\n\nExamples - a ball will keep rolling until its friction with the ground stops it. A rock will stay put unless someone transfers energy to it to move it.\n\n2nd - Force = Mass x Acceleration\n\nExamples - Some car has 5000 kg mass and 10 m/s^2 acceleration. Find its force.\nForce = 5000 x 10\nForce = 50,000 newtons\n\nSome truck has 250,000 newtons of force, and an acceleration of 5 m/s^2. Find its mass.\n250,000 = Mass x 5\nMass = 250,000/5 = 50,000 kg   \n\n3rd - For every action, there is an equal and opposite reaction. \n\nExample - you bounce a ball on the ground. The ball hits the ground, and compresses inward. The ground pushes back up on the ball causing it to land in your hands. You hit a volleyball in PE. The ball pushes against your hands and goes over the net.
512623
I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
733431
Because there is no absolute reference frame for motion, you can only define uniform motion with respect to a specified coordinate frame.\n\nFor example, a ball falling in a vacuum is not in uniform motion with respect to an earth-fixed reference frame; the ball accelerates.\n\nWhen the forces on an obejct sum to zero, that object is in uniform motion with respect to a reference frame that is non-accelerated. That means that if you compare gravity acceleration between the a frame fixed to the object, and the frame you are wondering about, if the gravity is the same, the frames are not accelerated with respect to each other.\n\nOthers have suggested that uniform motion requires absence of forces. This is not quite true - it requires absence of *net* forces. For example, a car traveling a constant velocity on a highway is not in a force-free environment: you can identify forces of air friction, propulsion from the wheels, weight, and support from the pavement. However it is considered to be in uniform motion with respect to a reference frame fixed to the earth, because all of those forces sum to zero, and because the apparent gravity in a frame fixed to the car is the same as in a frame fixed to the earth.
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
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.
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
905757
This is uniform motion, if it is going in a straight line at constant speed.\n\nIf there are no forces acting upon a body, and therefore no acceleration (remember F=ma) or change in direction, nor any change in angular momentum, the motion is uniform. If it were changing direction, accelerating or going round in circles then it is non uniform.
593602
The answer is to draw an outline of the box and sketch all forces acting upon it.\n\nThen, resolve those forces into their horizontal and vertical components.\n\nThe box is not required to accelerate in order for constant-velocity motion to occur.  So, it is simpler than F = ma.  Just use Sum (F horizontal) = 0, and sum (F vertical) = 0.\n\nSolution:\n\nSum of horizontal forces:\n\nF - 0.3 * N * cos(12) * N * sin (12) = 0\n\nSum of Vertical forces:\n\n- mass * gravity - 0.3 * N * sin(12) + N * cos(12) = 0\n\nBe careful with pounds as a unit of mass.  Better to convert to slugs by dividing by 32.17 ft/s^2 (gravity).\n\nFirst solve for N (the normal force perpendicular to the plane acting upward on the box) using the vertical force eqn:\n\n-(35/32.17) * (32.17) - 0.3 * N * sin(12) + N*cos(12) = 0\n\nThen plug N into the Horizontal Force sum equation to solve for F.\n\nWatch your units!
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)
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
883912
Let the banking angle be '$'.\nThe reaction force will be perpendicular to the banking surface.\nLet the reaction force be 'R'.\nLet 'm' be the mass of the car.\nvelocity of car (v)=27 mt/s\nradius(r)=60 mt.\n\nThe centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.\nR sin$= mv^2/r ---------------------------(1) \n\nThe weight of the car will be vertically downward. This weight will be balanced by Rcos$.\nRcos$=mg  ----------------(2)\n\nNow dividing eqn(1) by eqn(2)\n\nRsin$/Rcos$=mv^2/r /mg\n=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239\n\n$= tan(inverse)(1.239)\n\nsolve above tan(inverse)  function. and find the value of the banking angle
91916
a car trvelling in a straight line\na beam of light