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igor_vitrenko [27]
3 years ago
11

A 2.4-m-diameter merry-go-round with a mass of 270 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in

the same direction that it is turning, and jumps onto the outer edge. John’s mass is 34 kg . Part A Part complete What is the merry-go-round's angular speed, in rpm, after John jumps on?
Physics
1 answer:
juin [17]3 years ago
5 0

Answer:

23.98 rpm

Explanation:

d = diameter of merry-go-round = 2.4 m

r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m

m = mass of merry-go-round = 270 kg

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²

M = mass of john = 34 kg

Moment of inertia of merry-go-round and john together after jump is given as

  I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²

w = final angular speed

w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s

v = speed of john before jump

using conservation of angular momentum

Mvr + I w₀ = I' w

(34) (5) (1.2) + (194.4) (2.093) = (243.36) w

w = 2.51 rad/s

w = 23.98 rpm

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3 years ago
True or false?
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1. False

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\epsilon= -\frac{d\Phi }{dt}

where \epsilon is the emf induced and \frac{d\Phi}{dt} is the rate of change of magnetic flux through the coil.

Lenz's law is a consequence of the law of conservation of energy. In fact, we have:

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2. False

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where

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Answer:

Power =  50204 [watts]

Explanation:

We know that the power is defined by the following expression:

Power = Work/time

where:

Power [watts]

time [seconds]

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Work = Force * distance [Joules]

Force[Newtons]

distance[meters]

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disa [49]

Answer:

a = 0.01m/s²

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a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

6 0
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