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 19 ∘ from horizontal. The tension in the rope is 2300 N. The force of the sail on the rider is 30∘ from horizontal.What is the weight of the rider?
1 answer:
Answer:
W = 506.75 N
Explanation:
tension = 2300 N
Rider is towed at a constant speed means there no net force acting on the rider.
hence taking all the horizontal force and vertical force in consideration.
net horizontal force:
F cos 30° - T cos 19° = 0
F cos 30° = 2300 × cos 19°
F = 2511.12 N
net vertical force:
F sin 30° - T sin 19°- W = 0
W = F sin 30° - T sin 19°
W = 2511.12 sin 30° - 2300 sin 19°
W = 506.75 N
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A) The new amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please find the attached files for the solution
Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- So, τ = 3/2*m*r²*α (2)
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.