A boat leaves a dock at 6:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 7:00 PM How many minutes after 6:00 PM were the two boats closest together

22 minutes

Explanation:

Let t be the time after 6 PM.

D be the distance

During time t, the First boat travelling south = 20t km.

Meanwhile, the second  boat travelling east is= 15t km,

Then the second boat  started 15 km from the dock, so the  distance from the  dock is= (15 - 15t) km.

The distance D between the  two boats is is expressed as = (15 - 15t)^2 + (20t)^2 = D*2

Let differentiate with respect to time  and making  dD/dt to 0 (D minimization is important),

then  2(15 - 15t)(-15) + 2(20t)(20) = 0

Therefore t = 9/25 h or 0.36 h or 22 minutes

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

your answer is yx

hope this

~mia