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2 answers:
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Answer:
ms⁻¹
Explanation:
= diameter of merry-go-round = 4 m
= radius of merry-go-round =
=
= 2 m
= moment of inertia = 500 kgm²
= angular velocity of merry-go-round before ryan jumps = 2.0 rad/s
= angular velocity of merry-go-round after ryan jumps = 0 rad/s
= velocity of ryan before jumping onto the merry-go-round
= mass of ryan = 70 kg
Using conservation of angular momentum
ms⁻¹
Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.