Yes
Four times the amount of traction is needed
if a truck weighing 5000 lbs traveling 15 mph hits an object , what is the pressure at impact?
You need the weight of the vehicle to calculate the force
4 times (from v squared)
106 km/hr = 65.87 mph (rounded)
YES
Pretty sure this is true...let me guess, the drug and alcohol test? I'm taking it too.
4 times the impact. The formula is 1/2 mass times velocity squared.
Four times the amount of traction is needed
NO
Tornadoes are capable of producing extremely powerful winds. Every year there are dozens of tornadoes that impact houses with winds in excess of 150 mph. In rare cases winds may exceed 300 mph. Few structures can withstand a 150 mph wind, let alone a 300 mph wind, which carries four times the force. In addition, strong tornadoes lift objects into the air and hurl them at high speeds, adding to the destructive potential.
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.
no
true
It would be 130 mph
"100" The time it takes for the impact to occur is equal to that of a single car hitting a wall at 100mph (in other words the distance between the two objects decreases at the same rate), but if both cars have the same mass the energy dispersed is equal between both cars so the equation would be F=MV/2 F being the force of the crash, which is equal to the mass of the two vehicles times the velocity of the vehicles divided by 2. This means that the impact is the same as one car absorbing the force of a 50mph crash....so the answer is 50mph