A stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R 227 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?
What a stunning, amazing question!!! the only hint right here which you like is locate the equation for centripetal tension. the vehicle will take off whilst the stress is triumph over. do not ignore to transform the respond from m/s to km/h!!!!!! ought to take you a optimum of two minutes to get the respond from interpreting this.
If they have the correct scanner it will work, all OBDII ports are supposed to be standard
I am a heretic. Purists say there is no such thing as centrifugal force. There is, you just introduce a Rotating Frame of Reference. Because: A rotating reference frame can have advantages over an inertial reference frame.[15][4] Sometimes the calculations are simpler (an example is inertial circles), and sometimes the intuitive picture coincides more closely with the rotational frame (an example is sedimentation in a centrifuge). By treating the extra acceleration terms due to the rotation of the frame as if they were forces, subtracting them from the physical forces, it's possible to treat the second time derivative of position (relative to the rotating frame) as if it was the absolute acceleration. Thus the analysis using Newton's law can proceed as if the reference frame was inertial, provided the fictitious force terms are included in the sum of forces. For example, centrifugal force is used in the FAA pilot's manual in describing turns.[16] Other examples are such systems as planets, centrifuges, carousels, turning cars, spinning buckets, and rotating space stations. So draw the picture. Centrifugal Force up Gravity down Fdown M*9.8 m/s^2 Force up M *(v^2/r) M*9.8 M*(V^2/r) Cancel the M 9.8 V^2/227 GO
