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4 years ago

Compare positive and negative feedback mechanisms.

2 answers:

7 0

Positive feedback mechanisms is the types of feedback loop in which a change in a particular direction causes additional change in same direction. it is also called positive feedback loops or positive feedback systems.

Negative feedback mechanism tends to bring conditions within the body back to it's normal state. it is also called negative feedback loops or negative feedback systems.

<h2>Further Explanation</h2>

Positive feedback mechanisms are mostly used for variables that have the ability to amplify itself. Positive feedback can be very harmful in most cases, and can also contribute to normal function when used in limited fashion.

An example of positive feedback occurs in lactation, in which a nursing mother produces breast milk for her infant. During pregnancy, the levels of the hormone (Prolactin) that stimulates milk production increases, but progesterone prevents milk production.

However, the progesterone levels will drop soon as a baby is born and the placenta is released from the uterus and this will cause the breast milk to flow. As the baby sucks the breast, the breast is also stimulated.

This stimulation causes further release of prolactin and more milk is produced. This positive feedback makes it possible for the baby to have enough milk during feeding.

Negative feedback mechanisms is a type of feedback loop in which a change in a direction causes changes in opposite direction.

For example, negative feedback loop that involves glucagon and insulin can help maintain a normal glucose level.

If the glucose is very high, the body discharges insulin into the bloodstream and if low, the body releases glucagon. The glucagon causes the release of glucose from the body cell.

LEARN MORE:

positive and negative feedback brainly.com/question/12299473

KEYWORDS:

positive feedback mechanisms
negative feedback mechanisms
hormones
glucose
feedback loops

NeX [460]4 years ago

6 0

Answer:

In positive feedback mechanism, the output is fed back to the system which further increases the output. Hence, it is known as positive feedback because it amplifies the output.

In the negative feedback mechanism, the output is fed back to the system which further decreases the the output. Hence, it is known as negative feedback because it reduces the output.

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Accurate _______ is needed in a valid experiment.<br><br> Hurry, I need help.

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3 years ago

A student uses a stopwatch to measure the period of the pendulum of the Beverly clock in the corridor. His measurements are: (a)

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Answer:

Reading is close to (b) 13.44 which is the best estimate of the period

Associated error, $\Delta E =0.178 s$

Given:

$t_{a} = 13.54 s$

$t_{b} = 13.44 s$

$t_{c} = 13.89 s$

$t_{d} = 13.41 s$

$t_{e} = 13.17 s$

$t_{f} = 13.22 s$

Solution:

1.The best estimate of the period can be calculated by the mean of the measurements and the one closest to the mean is the best estimate of the measurement:

$Mean, \bar {x} = \fra{sum of all observations}{No. of observation}$

$Mean, \bar {x} = \frac{t_{a} + t_{b} + t_{c} +t_{d} + t_{e} + t_{f}}{6}$

$Mean, \bar {x} = \frac{13.54 + 13.44 + 13.89 + 13.41 + 13.17 + 13.22}{6}$

$Mean, \bar {x} = 13.445 s$

It is close to 13.44 s

2. Associated error is given by:

$\Delta E_{n} = |measured value - actual value|$

$\Delta E_{n} = |t_{n} - \bar {x}|$

where

n = a, b,......, e

Now,

$\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01$

$\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00$

$\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45$

$\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03$

$\Delta E_{e} = |t_{e} - \bar {x}| = |13.17 - 13.44| = 0.027$

$\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10$

Mean Absolute Error, $\Delta E = \frac{\Sigma E_{n}}{6}$

$\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}$

$\Delta E =0.178 s$

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3 years ago

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

kakasveta [241]

Answer:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

$Vector (OA) = 6i + 0j + 0k \\Vector (OB) = 0i + 3j + 2k \\Vector (OC) = 0i - 2j + 3k$

Next step is to find unit vectors $u_{AB} ,u_{AC}, u_{AD}, u_{AE}$ as follows:

$u_{AB} = \frac{vector(AB)}{magnitude(AB)} \\= \frac{OB - OA}{magnitude({vector(OB - OA))} }\\=\frac{-6i +3j+2k}{\sqrt{6^2 + 3^2+2^2} } \\\\=-0.857 i +0.429j+0.286k\\\\u_{AC} = \frac{vector(AC)}{magnitude(AC)} \\= \frac{OC - OA}{magnitude({vector(OC - OA))} }\\=\frac{-6i -2j-3k}{\sqrt{6^2 + 2^2+3^2} } \\\\=-0.857 i -0.286j+0.429k\\\\u_{AD} = +1i\\u_{AC} = -1k$

Using the diagram we find the corresponding vectors Forces:

$F_{AB} = F_{AB} i + F_{AB}j +F_{AB}k\\F_{AC} = F_{AC} i + F_{AC}j +F_{AC}k\\F_{AD} = F_{AD} i + F_{AD}j +F_{AD}k\\W = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\\F_{AB}. u_{AB} + F_{AC}.u_{AC} + F_{AD}.u_{AD} + W = 0\\(-0.857F_{AB}i + 0.429F_{AB}j +0.286F_{AB}k) + (-0.857F_{AC}i - 0.286F_{AC}j +0.429F_{AC}k) + (+1F_{AD} i) + (-160k) = 0$