Ship a is located 3.90 km north and 2.50 km east of ship b. ship a has a velocity of 21.0 km/h toward the south and ship b has a velocity of 40.0 km/h in a direction 37.0° north of east. what are the (a) x-component and (b) y-component of the velocity of a relative to b? (axis directions are determined by the unit vectors i and j, where i is toward the east.) (c) at what time is the separation between the ships least? (d) what is that least separation?

a) Vx = -31.95 km/h       b) Vy = -45.07 km/h

c) t = 0.083 h                 d) d = 0.22 km

Explanation:

First we have to express these values as vectors:

ra = (2.5, 3.9) km      rb = (0,0)km

Va = (0, - 21) km/h    Vb = (31.95, 24.07) km/h

Now we can calculate relative velocity:

For parts (c) and (d) we need the position of A relative to B and the module of the position will be de distance.

In order to find out the minimum distance we have to derive and find t where it equals zero:

Solving for t we find:

t = 0.083 h

Replacing this value into equation for d:

d = 0.22 km

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

radiation

explanation:

heat is transmitted by radiation