Respuesta :
Answer:
The velocity of the ship relative to the earth V = 9.05 [tex]\frac{m}{s}[/tex]
Explanation:
The local ocean current is = 1.52 m/s
Direction [tex]\theta[/tex] = 40°
Velocity component in X - direction [tex]V_{x}[/tex] = 1.52 [tex]\cos 40[/tex]°
[tex]V_{x}[/tex] = 1.164 [tex]\frac{m}{s}[/tex]
Velocity component in Y - direction [tex]V_{y}[/tex] = 8 + 1.52 [tex]\sin 40[/tex]°
[tex]V_{y}[/tex] = 8.97 [tex]\frac{m}{s}[/tex]
The velocity of the ship relative to the earth
[tex]V = \sqrt{V_{x}^{2} + V_{y} ^{2} }[/tex]
Put the values of [tex]V_{x}[/tex] and [tex]V_{y}[/tex] we get,
⇒ [tex]V = \sqrt{1.164^{2} + 8.97 ^{2} }[/tex]
⇒ V = 9.05 [tex]\frac{m}{s}[/tex]
This is the velocity of the ship relative to the earth.
Answer:
Second part of question on cengage: Earth in degrees north of east?= 82.61
Explanation:
tan-1= (8.97/1.164)
82.61
