a = F / m = (mg * sin(30.0°)) / m = g * sin(30.0°) = 9.80 m/s^2 * 0.500 = 4.90 m/s^2
θ = tan^(-1)(-30.3 km / 37.5 km) = -38.3°
Using Newton's second law:
North component: 20.0 km + 35.0 km * cos(60.0°) = 20.0 km + 17.5 km = 37.5 km West component: -35.0 km * sin(60.0°) = -30.3 km a = F / m = (mg * sin(30
:
So, the magnitude of the resultant displacement is 48.2 km, and its direction is 38.3° south of west.
A 3.00-kg block is pushed up a frictionless ramp that makes an angle of 30.0° with the horizontal. Find the block's acceleration. a = F / m = (mg * sin(30