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marishachu [46]
3 years ago
5

Verify that for values of n less than 8, the system goes to a stable equilibrium, but as n passes 8, the equilibrium point becom

es unstable, and a stable oscillation is created.
Physics
1 answer:
irakobra [83]3 years ago
5 0

Answer:

Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.

Explanation:

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As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate
Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current I. Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field B. The direction of the right thumb should now point in the direction of the force on the wire F.)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

7 0
3 years ago
A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?
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6 0
3 years ago
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The amount of water vapor in the air is known as the
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3 years ago
A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu
ira [324]

Answer:

-20.0 kg m^2/s

Explanation:

The angular momentum of an object in rotation is given by

L=I \omega

where

I is the moment of inertia

\omega is the angular speed

In this problem, initially we have

I=2 kg m^2 is the moment of inertia of the wheel

\omega_i = 6.0 rad/s is the initial angular speed

So the initial angular momentum is

L_i = I\omega_i = (2)(6.0)=12 kg m^2/s

Later, a counterclockwise torque of

\tau=-5.0 Nm is applied

So the angular acceleration of the wheel is:

\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2 in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

\omega_f = \omega_i + \alpha t

where

t = 4.0 is the time interval

Solving,

\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s

So, the change in angular momentum is:

\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2

7 0
3 years ago
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