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marin [14]
3 years ago
15

Suppose a disk with constant angular velocity has rotational kinetic energy 1280 J. If the moment of inertia of the disk is 35 k

g-m^2, then what is its angular velocity? (a) 7.604 rad/s (b) 8.552 rad/s (c) 10.12 rad/s (d) 6.818 rad/s (e) 9.952 rad/s (f) 8.935 rad/s f
Physics
1 answer:
lozanna [386]3 years ago
8 0

Answer:

Angular velocity of the disk is 8.552 rad/s

Explanation:

It is given that,

Rotational kinetic energy, KE = 1280 J

The moment of inertia of the disk, I = 35 kg m²

We have to find the angular velocity of the disk. In rotational mechanics the kinetic energy of the disk is given by :

KE=\dfrac{1}{2}I\omega^2

\omega=\sqrt{\dfrac{2KE}{I}}

\omega=\sqrt{\dfrac{2\times 1280\ J}{35\ kgm^2}}

\omega=8.552\ rad/s

Hence, the angular velocity of the disk is 8.552 rad/s.

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In 1977 off the coast of Australia, the fastest speed by a vessel on the water
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Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

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