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goldfiish [28.3K]
2 years ago
7

Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or

not at all.
Physics
1 answer:
slamgirl [31]2 years ago
5 0

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

To learn more about feedback loop, visit:

brainly.com/question/11312580

#SPJ4

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A scientist wants to study the effects of morning temperatures on flower width. Which is most likely a source of error in this s
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<span>The flower width data were collected at about the same time every day. </span>  

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4 years ago
The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it
Schach [20]

Answer:

The tub of the washer does 37.5 revolutions while it is in motion.

Explanation:

At first we assume that the tub accelerates and decelerates uniformly. In this case, it is required to determine the number of revolutions done by the tub by means of the following equations of motion:

Acceleration

\ddot n_{1} = \frac{\dot n_{1}-\dot n_{1,o}}{t_{1}} (Eq. 1)

Where:

\ddot n_{1} - Angular acceleration experimented by the tub, measured in revolutions per square second.

\dot n_{1,o}, \dot n_{1} - Initial and final angular velocities of the tub, measured in revolutions per second.

t_{1} - Acceleration time, measured in seconds.

\Delta n_{1} = \dot n_{1,o}\cdot t_{1}+\frac{1}{2}\cdot \ddot n_{1}\cdot t_{1}^{2} (Eq. 2)

Where \Delta n_{1} is the change in angular position of the tub during acceleration, measured in revolutions.

Deceleration

\ddot n_{2} = \frac{\dot n_{2}-\dot n_{1}}{t_{2}} (Eq. 3)

Where:

\ddot n_{2} - Angular acceleration experimented by the tub, measured in revolutions per square second.

\dot n_{1}, \dot n_{2} - Initial and final angular velocities of the tub, measured in revolutions per second.

t_{2} - Deceleration time, measured in seconds.

\Delta n_{2} = \dot n_{1}\cdot t_{2}+\frac{1}{2}\cdot \ddot n_{2}\cdot t_{2}^{2} (Eq. 4)

If we know that \dot n_{1,o} =0\,\frac{rev}{s}, \dot n_{1} = 5\,\frac{rev}{s}, t_{1} = 7\,s, \dot n_{2} = 0\,\frac{rev}{s} and t_{2} = 8\,s, then the amount of revolutions done by the tub during each phase is, respectively:

(Eq. 1)

\ddot n_{1} = \frac{5\,\frac{rev}{s}-0\,\frac{rev}{s}  }{7\,s}

\ddot n_{1} = \frac{5}{7}\,\frac{rev}{s^{2}}

(Eq. 3)

\ddot n_{2} = \frac{0\,\frac{rev}{s}-5\,\frac{rev}{s} }{8\,s}

\ddot n_{2} = -\frac{5}{8}\,\frac{rev}{s^{2}}

(Eq. 2)

\Delta n_{1} = \left(0\,\frac{rev}{s} \right)\cdot (7\,s)+\frac{1}{2}\cdot \left(\frac{5}{7}\,\frac{rev}{s^{2}}\right) \cdot (7\,s)^{2}

\Delta n_{1} = \frac{35}{2}\,rev

(Eq. 4)

\Delta n_{2} = \left(5\,\frac{rev}{s} \right)\cdot (8\,s)+\frac{1}{2}\cdot \left(-\frac{5}{8}\,\frac{rev}{s^{2}}\right) \cdot (8\,s)^{2}

\Delta n_{2} = 20\,rev

The total amount of revolutions done by the tub while it is in motion is:

\Delta n_{T} = \Delta n_{1}+\Delta n_{2}

\Delta n_{T} = 37.5\,rev

The tub of the washer does 37.5 revolutions while it is in motion.

4 0
3 years ago
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