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g100num [7]
3 years ago
15

A boat is travelling down stream in a river at 20 m/s south. A passenger on the boat walks toward the back of the boat at 10 m/s

north. Relative to the ground, at what velocity is the passenger moving?
A) 10 m/s north
B) 10 m/s south
C) 30 m/s north
D) 30 m/s south
Physics
1 answer:
artcher [175]3 years ago
8 0

The question is unclear regarding the boat's velocity. Is it 20 m/s south relative to the water, or relative to the earth? (It is a river, after all...)

There's also the possibility that the boat's velocity relative to the river is 0. Take the south direction to be negative and north to be positive, and denote by v_{A/B} the velocity of a body A relative to a body B. Under these conditions,

v_{B/E}=v_{B/W}+v_{W/E}\iff-20\,\dfrac{\mathrm m}{\mathrm s}=0+-20\,\dfrac{\mathrm m}{\mathrm s}

(B for boat, E for earth, W for water) so that the passenger's velocity relative to the earth is

v_{P/E}=v_{P/B}+v_{B/W}+v_{W/E}

(P for passenger)

v_{P/E}=10\,\dfrac{\mathrm m}{\mathrm s}+0-20\,\dfrac{\mathrm m}{\mathrm s}=-10\,\dfrac{\mathrm m}{\mathrm s}

or 10 m/s south.

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vladimir1956 [14]

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rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

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Final angular velocity; ω_2 = 19 rpm

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distance from the centre; r = 1.55 m

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Using the principle of conservation of angular momentum, we have;

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Iω_1 = I'ω_2

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37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

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Thus, the moment of inertia of the merry go round is 38.04 kg.m²

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