A package of mass 7 kg sits at the equator of an airless asteroid of mass 3.0 1020 kg and radius 5.7 105 m. We want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed 236 m/s. We have a large and powerful spring whose stiffness is 2.2 105 N/m. How much must we compress the spring?

Centrapedial force is the answer

Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.