You are on the west bank of a river which flows due south and want to swim to the east bank (draw a picture!). You have told your friends to meet you on the east bank directly opposite your starting point and there is no bridge so, being the nut you are, you decide to swim it. Before starting out, you realize that, since the river is flowing swiftly at a speed of 12 ft/s and since your fastest swimming speed in still water is only 5 ft/s, you will inevitably be carried downstream. Nevertheless, you want to minimize the effort expended by your friends in walking downstream to meet you. Your guide book to the region tells you that the width of the river is 300 ft. After a quick calculation, you call your friends on your cellular phone and tell them to start walking to a new meeting point. Required:
How far downstream of the original meeting point should you tell them to walk?

 θ = 50 ,  y' = 360 ft

Explanation:

This exercise is related to the addition of speeds, as they ask us the lower the starting point the swimmer reaches.

Let's solve the problem in parts, let's start by knowing the time it takes to reach the opposite shore, its velocity of v = 5ft / s and the width of the river is x = 300 ft

            v = d / t

            t = d / v _man

            t = 300/5

            t / 60 s

This is the time it takes to get to the opposite shore if there is no current, now let's find when the current of the river that goes to the South diverts it

           v_river = y / t

           y = v_river t

           y = 12 60

           y = 720 ft

To minimize the distance that your friend must travel, we can glass the distance, for this the swimmer must swim at an angle with respect to the river.

So let's use trigonometry to find out what angle you should swim at to cover

         y ’= y / 2

         y ’= 720/2 = 360 ft

         tg θ= y '/ x

         θ = tg⁻¹ y'/ x

         θ = tg⁻¹ 360/300

         θ = 50

This is the angle at which the friend must swim the minimum distance downstream.

option number c is the right answer.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j