Principles Of Physics A Calculus-based Text 5th Edition Solution < DIRECT >

A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north. Find the magnitude and direction of the car's resultant displacement.

Using Newton's second law:

Let's break down the displacement into its north and west components:

So, the magnitude of the resultant displacement is 48.2 km, and its direction is 38.3° south of west. A car travels 20

The resultant displacement is:

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(30.0°)) / m = g * sin(30.0°) = 9.80 m/s^2 * 0.500 = 4.90 m/s^2 The resultant displacement is: North component: 20

:

The direction is:

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 car travels 20

The force acting on the block is:

F = mg * sin(30.0°)