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Serggg [28]
3 years ago
6

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

t of 4.9 rad. What is its average angular acceleration
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
4 0

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

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The noble gases neon (atomic mass 20.1797 u) and krypton (atomic mass 83.798 u) are accidentally mixed in a vessel that has a te
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Answer:

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(b) v(rms) Ne= 659.6m/s and v(rms) Kr= 323.7m/s

Explanation:

(a) According to the kinetic theory of gases the average kinetic energy of the gases can be calculated by:

K_{av} = \frac{3}{2}kT (1)        

<em>where K_{av}: is the kinetic energy, k: Boltzmann constant = 1.38x10⁻²³J/K, and T: is the temperature </em>

<u>From equation (1), we can calculate the</u><u> average kinetic energies for the krypton and the neon:</u>

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(b) The rms speeds of the gases can be calculated by:

K_{av} = \frac{1}{2}mv_{rms}^{2} \rightarrow v_{rms} = \sqrt \frac{2K_{av}}{m}  

<em>where m: is the mass of the gases and v_{rms}: is the root mean square speed of the gases</em>

For the neon:

v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{20.1797 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 659.6 \frac{m}{s}          

For the krypton:

v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{83.798 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 323.7 \frac{m}{s}  

Have a nice day!

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