Some race tracks utilize banking turns to allow cars to taking the turns at much higher speeds. Below is a depiction of

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Some Race Tracks Utilize Banking Turns To Allow Cars To Taking The Turns At Much Higher Speeds Below Is A Depiction Of 1 (58.62 KiB) Viewed 11 times

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Some race tracks utilize banking turns to allow cars to taking the turns at much higher speeds. Below is a depiction of a car coming straight at you while on the banked turn. 35" Part A) If the mass of the car is 648kg what is the weight of the car? W=6350.4 unit N pick a direction from below no direction, it is a scalar no direction, the magnitude is zero

Part B) It turns out if the car wasn't moving the normal force on the car would be equal to a component of the force of gravity and there would be some necessary friction between the tires and the track to keep it from sliding down the embankment. But in our problem the car is currently moving around the turn and getting turned by the 3-d shape of the track, which actually makes the normal force larger than the weight of the car. Let's say the driver wanted to go around the turn at the perfect speed where the car wouldn't want to slide up or down the bank and they could go around in a perfect circle(he/she wouldn't need friction to maintain their height on the bank) What is the normal force on the car in this situation? Fn =5207.3 x unit N pick a direction from below no direction, it is a scalar no direction, the magnitude is zero

Part C) What is the net force on the car? (this will be equal to the component of the normal force that is pushing the car toward the center of the turn, making it travel in a circular path in the horizontal plane) Fnet = 3619.7 x unit N pick a direction from below ← 15175//\ no direction, it is a scalar no direction, the magnitude is zero Part D) The net force gives the car what centripetal acceleration? ac = 5.60 x unit pick a direction from below ↑ لام no direction, it is a scalar no direction, the magnitude is zero

Part E) If the radius of the turn was 152.3m then what is this perfect speed that the car would need to be going in order to not utilize friction? S = 29.20 x unit m 8 pick a direction from below X 1/ O no direction, it is a scalar no direction, the magnitude is zero Question Help: Message instructor Submit Question