In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
ider is towed at a constant speed by a rope that is at an angle of 15 ∘ from horizontal. The tension in the rope is 1900 N. The force of the sail on the rider is 30∘ from horizontal. What is the weight of the rider? Express your answer with the appropriate units.
1 answer:
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
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Answer:
Explanation:
To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:
(1)
At the beginning the girl is stationary:
(2)
If the girl catch the ball, both have the same speed:
(3)
We replace (2) and (3) in (1):
We can now solve the equation for v_{f}:
Answer:
The initial velocity of the water from the tank is 5.42 m/s
Explanation:
By applying Bernoulli equation between point 1 and 2
At the point 1
P₁=0 ( Gauge pressure)
V₁= 0 m/s
Z₁=3 m
At point 2
P₂=0 ( Gauge pressure)
Z₂= 0 m/s
Now by putting the values
V₂= 5.42 m/s
The initial velocity of the water from the tank is 5.42 m/s
