Answer:
a)
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
the required distance is 40.98 m
Explanation:
Given that;
velocity of the river u = 1.70 m/s
velocity of boat v = 14.0 m/s
Now to get the velocity of the boat relative to shore;
( north of east), we say
a² + b² = c²
(1.70)² + (14.0)² = c²
2.89 + 196 = c²
198.89 = c²
c = √198.89
c = 14.1028 m/s
tan∅ = v/u = 14 / 1.7 = 8.23529
∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east
Therefore, the velocity of the boat relative to shore is;
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
width of river = 340 m,
ow far downstream has the boat moved by the time it reaches the north shore in meters = ?
we say;
340sin( 90° - 83.0765°)
⇒ 340sin( 6.9235°)
= 40.98 m
Therefore, the required distance is 40.98 m
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. The criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the <span>particles remain suspended for an extended period of time. I hope it helps you.</span>
C) light waves travel faster than sound waves
You were correct
Answer:what are the answer options?
Explanation
Answer:
0.687 m/s
Explanation:
Initial energy = final energy
1/2 mu² = mgh + 1/2 mv²
1/2 u² = gh + 1/2 v²
Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:
1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²
v = 0.687 m/s
