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svet-max [94.6K]
3 years ago
6

Determine the total moment of inertia of a merry-go round with 5 children sitting on it. Of the five children, four are seated a

t the edge of the merry-go-round, and the fifth child is sitting exactly at the center. The children have the same mass of 30.0 kg. You may consider the merry-go-round as a cylinder of mass 150 kg and a radius 2.0 meters.
Physics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

Explanation:

Given that,

We have five children.

Each of mass m =30kg

They sit on a merry go round

Mass of Merry go round M= 150kg

Radius of Merry go round is r =2m

Four children sit at the edge of the merry go round but one child sit at the centre.

The four child that sit at the edge are 2m from the centre of the merry go round but the one at the centre is 0m from the centre

Moment of inertia?

Moment of inertia is given as

I=Σmi•ri²

For the question, the moment of inertia is the combination of inertial of child and the merry go round

I= I(merry go round) + I(four child)+ I(last child)

The merry go round is assumed to be a solid cylinder, so it is going to have the moment of inertia of solid cylinder

Then,

I(merry go round ) =½ Mr²

Also, Four of the child has the same moment of inertia, they are 2m form the centre of the merry go round why the last child has no moment of inertia

I= I(merry go round) + I(four child) +I(last child )

I= ½Mr² + 4mr² + mr'²

I = ½ × 150 ×2² + 4×30×2² + 30×0²

I = 300 +480+0

I = 780 kgm²

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Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}

Here,

\Delta L= is the distance moved by the mirror M

\lambda is the wavelenght of the light used.

\Delta L= 0.017m

\lambda = 656.45*10^{-9}m

\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}

\Delta m = 51793.72

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

3 0
3 years ago
A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is
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Answer:

a=16.2m/s^{2}

Explanation:

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When fixed,F=o

Hence

masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}

The value of the acceleration is 16.2m/s^2

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Acceleration is measured in meters per second square.
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Answer:

2

Explanation:

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Identify the velocity-versus-time plot(s) that correspond to motion under a constant, non-zero acceleration.
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Answer:

See explanation

Explanation:

The question is incomplete because the images were not attached but I will try to help you as much as possible.

Constant acceleration implies that the velocity increases uniformly with time.

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The image attached shows a velocity-time graph depicting constant acceleration.

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