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KengaRu [80]
3 years ago
5

The area of the bar over r = 2 is 0.234. what is the area of the bar over r = 4?

Physics
1 answer:
natita [175]3 years ago
4 0
The area of the bar over r=4 is 0.468
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What motion makes the acceleration unchanged?
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The acceleration is equal to the net force divided by the mass. If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

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A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp
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Answer:

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Explanation:

We can solve the problem by applying Newton's second law of motion, which  states that:

\sum F=ma

where:

\sum F is the net force acting on an object

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a is its acceleration

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in the same direction as the force (left).

5 0
3 years ago
At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

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

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

Where:

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

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

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

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

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

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

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

The angular acceleration is now cleared:

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

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

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

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

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
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