A massless spring is attached to a support at one end and has a 2.0 μC charge glued to the other end. A -4.0 μC charge is slowly brought near. The spring has stretched 1.2 cm when the charges are 2.4 cm apart. What is the spring constant of the spring?

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.