a rocket burns propellant at a rate of dm/dt = 3.0 kg/s, ejecting gases with a speed of 8000 m/s relative to the rocket. Find the magnitude of the thrust.

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