Arowboat crosses a river with a velocity of 3.30 mi/h at an angle 62.5â° north of west relative to the water. the river is 0.505 mi wide and carries an eastward current of 1.25 mi/h. how far upstream is the boat when it reaches the opposite shore?

 Rowboat arrives 0.558 miles to the upstream when it reaches the opposite shore.

Explanation:

   Velocity of rowboat = 3.30 mi/h

   Angle with north of west = 62.5⁰

   Width of river = 0.505 mile.

   Velocity of river = 1.25 mi/hr eastward.

Let us take east as positive X -axis and North as positive Y-axis.

   So angle of boat with horizontal axis = (90+62.5) = 152.5⁰

  Horizontal speed of boat = 3.30*cos 152.5 = -2.93 mi/h

  Vertical speed of boat = 3.30*sin 152.5 = 1.52 mi/h

  Horizontal speed of river = 1.25 mi/h

  Time taken to cross river = Width of river/Vertical speed = 0.505/1.52 = 0.33 hour.

  Velocity along upstream of river = -2.93+1.25 = -1.68 mi/hr

  So distance moved along upstream = -1.68*0.33 = -0.558 miles ( negative sign means along west direction)

  It reaches 0.558 miles to the upstream when it reaches the opposite shore.

continuous conducting path for current, power supply

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j