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3 years ago

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Two particles, one with charge − 7.97 μC and one with charge 3.55 μC, are 6.59 cm apart. What is the magnitude of the force that

one particle exerts on the other?

1 answer:

Arlecino [84]3 years ago

7 0

Answer:

58.6 N

Explanation:

We are given that

$q_1=-7.97\mu C=-7.97\times 10^{-6} C$

$q_2=3.55\mu C=3.55\times 10^{-6} C$

Using $1\mu C=10^{-6} C$

$r=6.59 cm=6.59\times 10^{-2} m$

$1 cm=10^{-2} m$

The magnitude of force that one particle exerts on the other

$F=\frac{kq_1q_2}{r^2}$

Where $k=9\times 10^9$

Substitute the values

$F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}$

F=58.6 N

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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

PilotLPTM [1.2K]

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:

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Where:

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

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

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

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:

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

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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:

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

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

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

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