Wo thin uniformly charged rods, each with length L and total charge +Q, are parallel and separated by a distance a. The first rod has one end at the origin and its other end on the positive y-axis. The second rod has its lower end on the positive x-axis Determine the x-component of the differential force dF2 exerted on a small portion of the second rod, with length dy2 and position y2, by the first rod. (This requires integrating over differential portions of the first rod, parameterized by dy1.)

he . if i may ask what i your question exactly?

explanation:

he made 11 calls in one day

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west