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zalisa [80]
3 years ago
14

A 120 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d

one on the hoop to stop it?
Physics
1 answer:
jeka57 [31]3 years ago
5 0

Answer:W = -5.808J

Explanation:

Given:

Mass of hoop,m=120kg

Centre of mass speed of hoop,v= 0.220m/s

Rotational inertia,I= mr^2

I=120r^2

Kinetic energy of hoop= Linear kinetic energy + Rotational kinetic energy

K= (1/2mv^2) + (1/2Iw^2)

But w= (v/r)^2

K= (1/2mv^2) + (1/2×120r^2(v/r)^2)

K = (1/2×120×0.220^2) + (1/2×120×0.220^2)

K= 5.808J

To stop the hoop,its final kinetic energy must be zero,

Workdone = Kfinal-Kinitial = 0 - 5.808

W= -5.808J

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Explanation:

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Explanation:

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Here

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  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

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      Here

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              f is the frequency of the sound which is to be calculated.

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where k=0,1,2

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So the minimum frequency is 702.22 Hz

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