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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Answer:
The time required is 20 μs
Explanation:
Here is the data provided:
temperature = 20°C = 293 K
radius = 1 cm
atomic mass of air = 29 u
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the duration for air to refill the vacuum space
solution:
We calculate the root mean square velocity of air particles. This can be expressed as:
where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.
v = ............................1
v =
Resulting in v = 501.99 m/s.
<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:time taken =
which gives us:
time taken =
so, time taken = 19.92 × seconds = 20μs.
Thus, the required time is 20 μs.