A man walks 1.80 km north and then 2.75 km east, all in 3.00 hours.(a)What is the magnitude (in km) and direction (in degrees north of east) of his displacement during the given time?magnitude km direction ° north of east(b)What is the magnitude (in km/h) and direction (in degrees north of east) of his average velocity during the given time?magnitude km/h direction ° north of east(c)What was his average speed (in km/h) during the same time interval?km/h

1. Displacement = 3.29 km

b. His direction is north of east

2. Average velocity is 1.10 km/h

b. His direction is north of east

3. Average speed = 1.52 km/h

Explanation:

1. Displacement =

                       =

                       =

                       = 3.2867

Displacement = 3.29 km

b. tan θ =

        θ =

           = 0.6546

           = 33.20878

 θ =

His direction is north of east

2. Average velocity =

                                =

                                = 1.09667

Average velocity is 1.10 km/h

b. His direction is north of east.

3. average speed =

total distance covered = 1.80 km + 2.75 km

                                      = 4.55 km

Average speed =

                          = 1.5167

Average speed = 1.52 km/h

pretty sure it's false

higher potential

explanation:

because it is in a high speed so it must move into a of high potential