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4 times the impact. The formula is 1/2 mass times velocity squared.
Without knowing the surface area of the moving object, there is not enough information to answer this question. Rephrase and resubmit.
To find the speed you would need to find the time taken to create the skid marks and then use the equations of motion. As the vehicle is skidding, the wheels are not turning so the braking force is the friction between the tyre and the road surface; this has to overcome the forward speed of the vehicle and the forward force of the weight of the vehicle along the road (as there is a downward slope). The forward force of the vehicle can be calculated by knowledge of the mass of the vehicle and the angle of the slope. The frictional force generated can be calculated from the coefficient of friction between the tyres and the road surface, and the mass of the vehicle. The road being wet will reduce the coefficient of friction of the road surface/tyre boundary compared to when the road surface is dry. It is easier, by experiment, to do a few test runs using a vehicle of similar mass under similar conditions to get how the vehicle is likely to have slowed down when it skidded, and extrapolate/interpolate to the given distance of 24.2 m.
The best way is to find the centre of surface of planar area. Then the force due to hydrostatic pressure will be:F = d h0 g S,where:F is force,d is density of fluid,h0 is depth at the centre of surface,S is surface of the area.It works because when we consider the centre of surface, there will exactly as much surface with lesser pressure effecting on it as there is surface below the centre point where the pressure is higher.The net force vector will be perpendicular to the area at the centre of surface point.
your cired
4 times.
Double the speed yeilds double the impact force.
Yes, all things being equal, crash severity does increase proportional to the speed of each vehicle at impact, and is a vector sum. So, there is a big difference between crash severity at impact from being "rear-ended" (when one vehicle is traveling the same direction as another, and impacts the front of their vehicle with the rear of another) and a "head-on" impact (two cars traveling into one another, impacting both front bumpers). In the rear-end impact, you take the momentum (mass times velocity) of the rear, impacting vehicle "A" and subtract the momentum of the front-most impacted vehicle "B", and that gives you the resultant impact force (the difference in momentum being transferred). weak impact scenario example: vehicle A is traveling 60 mph, and vehicle B is the same mass and is traveling 50 mph. The difference in momentum would be the mass times 10 mph...not much. severe impact scenario: vehicle A is traveling 70 mph, and vehicle B is at rest (0 mph)...large impact. In the head-on impact, you have the most severe crash scenario. In this case, you ADD the momentum of vehicle A with the momentum of vehicle B, and you get the resultant force of impact. Even if both vehicles are traveling 30 mph, with the same mass, and have a heaad-on collision, the is close to the same as one vehicle traveling 10 mph and hitting the other vehicle going 70 mph...severe impact.
Yes, all things being equal, crash severity does increase proportional to the speed of each vehicle at impact, and is a vector sum. So, there is a big difference between crash severity at impact from being "rear-ended" (when one vehicle is traveling the same direction as another, and impacts the front of their vehicle with the rear of another) and a "head-on" impact (two cars traveling into one another, impacting both front bumpers). In the rear-end impact, you take the momentum (mass times velocity) of the rear, impacting vehicle "A" and subtract the momentum of the front-most impacted vehicle "B", and that gives you the resultant impact force (the difference in momentum being transferred). weak impact scenario example: vehicle A is traveling 60 mph, and vehicle B is the same mass and is traveling 50 mph. The difference in momentum would be the mass times 10 mph...not much. severe impact scenario: vehicle A is traveling 70 mph, and vehicle B is at rest (0 mph)...large impact. In the head-on impact, you have the most severe crash scenario. In this case, you ADD the momentum of vehicle A with the momentum of vehicle B, and you get the resultant force of impact. Even if both vehicles are traveling 30 mph, with the same mass, and have a heaad-on collision, the is close to the same as one vehicle traveling 10 mph and hitting the other vehicle going 70 mph...severe impact.
Yes. (1/2 mass x velocity squared)
Depends on the weight of the vehicle.
the force applied by the drive wheels is equal and opposite to the forces which slow the vehicle
force of impact basiclly means that a force is impacting something! ~hope it helped(:
4 times the impact. The formula is 1/2 mass times velocity squared.
double
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a sports car and a bus are both traveling at 30 km/h. which of the two will require more force to stop?why?