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gizmo_the_mogwai [7]
3 years ago
7

What is the difference between a negative feedback system and a positive feedback system?

Physics
1 answer:
Anni [7]3 years ago
7 0

Answer:

D

Explanation:

The negative feedback is responsible for maintaining equilibrium (stability) in a system as it lessens effects, which is contrary to positive feedback which increases reaction and moves a system further away from equilibrium (stability), It does so by amplifying the effects of a product or event and occurs when something needs to happen quickly. e.g

  • Insulin lowers down blood sugar levels, so when the body detects that it has too much sugar, the pancreas is prompted to release insulin and only stops when balance is achieved; hence, negative feedback.  
  •  When there is a tear on the skin, a chemical is released. This chemical causes platelets in the blood to activate, hence they release a chemical which signals more platelets to activate, until the wound is clotted, positive feedback.

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A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.
mars1129 [50]

Answer:

a) v = 31.67 cm / s , b)   v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

       w = √ K / m

       w = √ 23.2 / 0.37

       w = 7,918 rad / s

a) the expression against the movement is

        x = A cos (wt + Ф)

Speed ​​is

        v = dx / dt = - A w sin (wt + Ф)

 The maximum speed occurs for cos = ± 1

        v = A w

        v = 4.0 7,918

        v = 31.67 cm / s

b) as the object is released from rest

        0 = -A w sin (0+ Фi)

        sin Ф = 0

         Ф = 0

The equation is

        x = 4.0 cos (7,918 t)

        v = -4.0 7,918 sin (7,918 t)

        v = - 31.67 sin (7.918t)

     

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

          7,918 t = cos⁻¹ 1.5 / 4.0

          t = 1,186 / 7,918

          t = 0.1498 s

We look for speed

         v = -31.67 sin (7,918 0.1498)

         v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

         v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

         31.67 / 2 = 31.67 sin ( 7,918 t)

          7.918t = sin⁻¹ 0.5

         t = 0.5236 / 7.918

         t = 0.06613

With this time let's look for displacement

         x = 4.0 cos (7,918 0.06613)

        x = 3.46 cm

6 0
3 years ago
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