Given that the hyperbola has a center at (0,0), and its vertices and foci are on y-axis. This, the equation of the hyperbola is of the form
x²/a²-y²/b²=-1 (a>0, b>0)
In the equation, vertices are (0, +/-b) .
Thus,
b=60
Foci (0,+/-√(a²+b²))
thus
√(a²+60²)=65
hence solving for a²
a²=65²-60²
a²=625
a²=25²
hence the equation is:
x²/25²-y²/60²=-1
Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
Answer:
HOLAAAAAO
Step-by-step explanation:
