Would you feel heavier at the top of the ferris wheel or the bottom and why.
At the top of the ferris wheel: ma=mg-F F=mg-ma In this case the positive direction is down and the negative direction is down. At the bottom of the ferris wheel: m(-a)=mg-F F=mg+ma Since the ferris wheel is in circular motion, there is centripetal force and acceleration. The acceleration at the top of the wheel is downwards towards the center and at the bottom it is towards the top (in opposite directions).
Describe the field and charge movements of a conducting sphere with charge 4C inside a concentric insulating sphere with equal charge distribution of charge -2C inside a concentric conducting sphere of charge 3C.
In the innermost sphere, the 4C would be distributed to the inside surface of the sphere. The insulator would maintain its charge distribution. In the outermost shell, a charge of -2C would be distributed on the inner surface (the part closest to the insulator) and a charge of 5C would be distributed at the outer edge, giving the sphere a total charge of 5C. There would be no field in the sphere until the shell of the first conductor containing the 4C charge. There would be field inside the entire insulator. There would again be no field after the insulator layer until the shell where the 5C are.
If a function f(x) differentiable and continuous on [-4,4] where f(-4)=f(4), What kind of critical point must be present on this interval if f'x=25x-x^3, and at what value of x does this occur? Can you find the value of this max/min? If yes, find it. If no, why not?
Because of Rolle's theorem, we know that if on a continuous and differentiable function, if f(a)=f(b) there must be a point f(c) that has an instantaneous rate of change equal to the average rate of change [f(a)-f(b)]/(a-b) which is 0 since f(a)=f(b). So, we plug in 0 for f'(x) and get that 0=2x-4x^3=x(25-x^2). Therefore, there is a critical value at x=0, x=-5, and x=5. x=5 is the only value of x in the interval. To find whether this is a max or a min, we must perform the second derivative test. f''x=25-3x^2. f''(5)=25-3(5)^2=-50. Since f''(x)<0, at x=5 there is a maximum. There is no way to determine the y value of this maximum since there is no way to determine the integration constant.