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3 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]3 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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The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad

kenny6666 [7]

Answer:

a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,

c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

k = 2π / λ

λ = 2π / k

λ = 2π / 1.99 10⁷

λ = 3,157 10⁻⁷ m

w = 2π f

f = w / 2 π

f = 5.97 10¹⁵ / 2π

f = 9.50 10¹⁴ Hz

the wave speed is

v = 3,157 10⁻⁷ 9.50 10¹⁴

v = 2,9992 10⁸ m / s

b) The electric field is

Eo = 375 V / m

to find the magnetic field we use

E / B = c

B = E / c

B = 375 / 2,9992 10⁸

B = 1.25 10⁻⁶ T

c) The period is

T = 1 / f

T = 1 / 9.50 10¹⁴

T = 1.05 10⁻¹⁵ s

the wavelength value is

λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

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3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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