Three people, each with a mass of 71.1 kg, are in a car with a mass of 1130 kg. An earthquake strikes. The driver manages to pull of the road and stop, as the vertical oscillations of the ground surface make the car bounce up and down on its suspension springs. When the frequency of the shaking is 1.60 Hz, the car exhibits a maximum amplitude of vibration. The earthquake ends and the three people leave the car as fast as they can. By what distance does the car‘s undamaged suspension lift the car‘s body as the people get out?(give answer in cm)Im pretty sure you have to use these equations: *Let mass of all 3 peoplem *Let mass of carM(2*Pi*f)^2 k/(M+m)k*delta(x) (M+m)*gThe aim is to get delta(x) which is the amplitude in relation to the equilibrium position.I got 9.70cm to 3 s.f. (0.09696753093 m) and it said it was wrong :(
find the spring constant of the car's suspension. m total 1343.1 kg and use f 1/(2pi)√(k/m) (1.6*2pi)^2*1343.1 k k 135 740 N/m Now 3 persons with 3*71.1 kg are a force of 2090 N and the distance by which this force compresses the spring is 2090/135 740 0.01539 m The car will lift by 0.01539. m when the 3 persons leave the car OG
It's just like anything else in life. The more $$$ you spend, usually the better the product. Continental tires (the more expensive ones) are also good. Better resistance to cuts flats.
It's just like anything else in life. The more $$$ you spend, usually the better the product. Continental tires (the more expensive ones) are also good. Better resistance to cuts flats.
find the spring constant of the car's suspension. m total 1343.1 kg and use f 1/(2pi)√(k/m) (1.6*2pi)^2*1343.1 k k 135 740 N/m Now 3 persons with 3*71.1 kg are a force of 2090 N and the distance by which this force compresses the spring is 2090/135 740 0.01539 m The car will lift by 0.01539. m when the 3 persons leave the car OG
