(2) Two masses are connected by a light string and hung over a light, frictionless pulley (meaning we can neglect the masses of the string and pulley, and any friction forces in the equations). We will specify that the positive x-direction is to the left over the pulley as shown. Thus motion in the positive direction means that m, drops while m2 rises. (Now consider the opposite – what motion in the negative direction would mean.) Consider the system of masses at left. + + m

Newton's universal law of gravitation can be stated as the force F of gravitation between two objects varies jointly as the masses m, and m, of the objects, and Gm,m2 inversely as the square of the distance r between their centers, where G is the constant of proportionality. In other words, F = The weight of an object is the force F on the object due to gravity, where F = mg. State how g (the acceleration due to gravity) varies with respect to the mass m, of the spherical body on which the object lies, and the radius of that body (assume all of the mass is at its center.). The acceleration due to gravity, g, varies as the spherical body's mass, and as

Solve the following problem (NOTE: Use any reasonable DATA not given in the problem.)....

Problem 2. 1-3

Consider a water tank that is being filled with rainfall at the top and is drained through a small hole atthe bottom. The change in volume can be modelled by considering the amount of water entering andleaving the tank (per unit of time) as followsdVdt= fin − fout.Here volume of water in the tank V is in litres, flow rate in fin is in litres per hour, flow rate out fout isin litres per hour and time t is in hours. Initially, the tank contains 50L of water: V (0) = 50.(a) Assume the rainfall throughout the day is getting heavier, such that, fin(t) = 10 + t andthat the tank is losing 10% of its volume of water per hour, such that, fout(V ) = 0.1V . Check bydirect substitution that the functionV (t) = 10(t + 5e−0.1t)satisfies the ODE and the initial condition.Now assume that the hole is slowly growing in size, so that the flow rate out increaseswith time:fout(V, t) = 0.1(1 + 0.1t)V.Solve the ordinary differential equation for V (t) with this flow rate out, using either separation…