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:
To analyze the forces acting on the rider, begin by sketching a free body diagram. Three forces can be identified: the tension force at 15° below horizontal, the drag force at 30° above horizontal, and the weight directed downward.
As the rider maintains a constant speed, the acceleration remains at 0.
Considering forces in the x-direction:
∑F = ma
F cos 30° - T cos 15° = 0
Thus, F = T cos 15° / cos 30°
Considering forces in the y-direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
From which, W = F sin 30° - T sin 15°
Substituting gives:
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's weight is 570 N, which is roughly equivalent to 130 lb.
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Respuesta:
11.4 m/s
Explicación:
La fórmula para la aceleración centrípeta es:
donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.
En este problema,
a = g = aceleración debida a la gravedad en la cima =
v = ?
R = 13.2 m
Por lo tanto,
v = 11.4 m/s
Answer:
The pressure measured at this moment is 0.875 mPa
Explanation:
Given that,
Flow energy = 124 L/min
Boundary to system P= 108.5 kJ/min
We are tasked with finding the pressure here
Applying the pressure formula
Here,
Where, v refers to velocity
Insert the values into the equation
Therefore, the pressure at this moment is 0.875 mPa