What net force acted on this car?
109202
Fnet= MA so you acceleration is 36km/h, so convert that to m/s, which is 10 m/s. To get acceleration, we know that the car started out at 18km/s, which is 5m/s. So the acceleartion = Change of velocity/Change of time. So that comes out 1 m/s/s. So there is 1000 Newtons on the car. The other guy's answer is wrong. I'm an engineering major. The standard Units in science is MKS, meters kilograms and seconds
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.
363865
As the car moves along, air passes over the top of the roof, creating an area of low pressure.  The air pressure inside the car, having a higher pressure than outside, then forces the canvas upward.\n\nThe principle is similar to that of an airplane wing.  Of course though, there isn't anywhere enough lift generated to make the car fly.
1022513
You probably mean "identify the frictional force acting on this object." Since we don't know what the object is, we can't really help you...
816243
When an object (like the truck) hits a solid wall then the wall forces it to stop suddenly. \n  The wall and the truck have a certain elasticity, though small. When the truck touches the wall it starts pressing on it and the wall bends very little. It bend in a very short time too. During this small period of time the object's momentum changes from "mv" to "0". The force is the ratio F= (m*v) / t. Because t is very small the force exerted is very big and it crushes the truck's front end.
905447
The net force is 0 Newtons. The force of friction is 100N.\n\nThe key here is that\n1. No acceleration.\n2. Horizontal force only.\n\nThis means that the forces are balanced in the horizontal direction. The object isn't accelerating so net F = ma = 0.
159509
Start with a toy car sitting on a table.  Point out that the car is not moving. Point out that no unbalanced external forces act on the car, hence it remains at rest, i.e. doesn't accelerate.  This demonstrates Newton's First law.  You can also demonstrate the other part of Newton's First Law, about objects in motion staying in motion unless acted upon by an outside force, by pushing the car and catching it a short distance later, before the force of friction had much opportunity to act on it.  Or you could use a marble, which is less affected by friction.\n\nFor Newton's Second law, use two cars, one more massive than the other.  You can arrange a weight and pulley sytem to pull the cars along if you want, or you could just tell your audience about the forces you are using as you push the cars.  Push 1 car lightly and see how fast it goes.  Then push it harder and see if it goes faster.  Next, push a more massive car and a less massive car with the same force, and see which ends up going faster.  The cars that went faster were accelerated more as they were pushed.  Thus, you "showed" that the acceleration of an object is directly related to the force accelerating it, and inversely related to the mass of the object.  This yields Force is proportional to mass x acceleration, (Newton's Second Law)\n\nFor the Third Law, have your audience imagine what it feels like to push the car.  It is impossible to push on something without it pushing back on you.  And the amount of the push-back is equal to the amount you push.  If they slam their hand against the car very hard, they will feel more force applied to their hand than if they gently push it.  And of course the push-back is in the opposite direction of the push.  If they push against the car to the right, the car pushes back to the left.  Hence, for every action, i.e. applied force,  there is an equal and opposite reaction force.
368233
N is Newtons, a unit that measures force. For every 1kg of something, if it accelerates at 1 meter per second per second (m/s/s or meters per seconds squared), than there is a force of 1N acting on it.\n\nForce is calculated by multiplying the mass (not weight) with the acceleration. For example, if you are 75kg, than there is a force of 75kg X 9.8 m/s/s (acceleration due to gravity: roughly constant on Earth) = 735N pulling you down. BUT, there is a reaction force, in this case the NORMAL, pushing you back up at the same time and force. Therefore, the force pulling you down, 735N, take away the NORMAL, also 735N, and you will have a NET FORCE of 0N because you are stationary. \n\nSo in order to resist something pulling at 14N (ignoring air resistance, friction, etc), it needs to pull back 14N.
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)
641452
opposite force being applied
572487
Distribution of force over area.
935603
The net force is F2 - F1 = 25 - 19 = 6N.\n\nFrom F = ma, a = F/m = 6/5 = 1.2 m/s^2, toward the west because F2 > F1.  Also, we're ignoring friction.
256930
dba_gregg is correct.\n\nThe gravity pulls the front while the back is still supported by normal reaction force. Thus a torque is acting so as to rotate the bus with the front down. The angular momentum produced is conserved and the bus begins to rotate about the free axis passing through its centre of mass.
845823
the speeds are too small, so classical mechanics should be fine:\n\nthe speed of the car is 45-10 = 35mph
490009
Because the normal force of the road upon the car is not directed straight upwards, as it is on a flat road. The normal force vector is tilted inward toward the center of the radius of curvature (assuming that the road engineers built the darned thing correctly) by an amount equal to the angle of the bank of the road. Decomposing the normal force vector into a horizontal (towards the center of the circle) and a vertical (up towards the sky) shows an increased force upon the car towards the center of the radius of curvature over the case where the road is flat. In the end, it means that the coefficient of static friction that keeps the wheels from sliding radially away from the center of curvature doesn't need to be as high when the road is banked.\n\nAlso note: There is NO SUCH THING AS CENTRIFUGAL FORCE! The car is an accelerating reference frame. Anyone who uses the term "centrifugal force" either has not taken 1st year physics or did not understand it when they did.
1163276
The angle the vehicle was traveling
784877
How are they applied? Tangentially? In opposite direction? Or same?\nLet us assume oppo direc and on the border.\nThen net force=35N.\nTorque = r x F = I w\n=35x0.341= ''0.5'' m R^2\nR radius m mass\n ''Check if it is 0.5 only for a solid disk or cyliner about its axis.\n\nHope it helps