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grandymaker [24]
4 years ago
7

An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.32 kg * m2.

Physics
1 answer:
ch4aika [34]4 years ago
4 0

Explanation:

The angular momentum is given by the moment of inertia, multiplied by the angular speed of the rotating body:

L=I\omega

The angular speed is given by:

\omega=2\pi f\\\omega=2\pi 5.2\frac{rev}{s}\\\omega=32.67\frac{rad}{s}

Now, we calculate the angular momentum:

L=0.32kg\cdot m^2(32.67\frac{rad}{s})\\L=10.45\frac{kg\cdot m^2}{s}

The average torque is defined as:

\tau=I\alpha

\alpha is the angular acceleration, which is defined as:

\alpha=\frac{\omega_f-\omega_0}{t}

We have to calculate \omega_f:

\omega_f=2\pi (2.75\frac{rad}{s})\\\omega_f=17.28\frac{rad}{s}

Now, we calculate the angular acceleration:

\alpha=\frac{17.28\frac{rad}{s}-32.67\frac{rad}{s}}{12s}\\\alpha=-1.28\frac{rad}{s^2}

Finally, we can know the average torque:

\tau=0.32kg\cdot m^2(-1.28\frac{rad}{s^2})\\\tau=-0.41N\cdot m

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AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
4 years ago
What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
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Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

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now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

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F_[net} = 150 N

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The answer cannot be determined.

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The change in temperature {\Delta}T then will be

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