In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50×10^6 m/s. The proton comes momentarily to rest at a distance 5.31×10^(−13)m from the center of the target nucleus, then flies back in the direction from which it came. What is the number of the protons the nucleus has? Assume no electron cloud is there, ε0=8.85x10^(-12) C^2/(Nm^2)

Ithink your answer is d

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction 

Since the pully is a simple machine so the mechanical advantage is a 1