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lidiya [134]
3 years ago
6

Which is a feedback mechanism for a system?

Engineering
1 answer:
stira [4]3 years ago
6 0

The following is a feedback mechanism for a system :

<u>The progress bar when downloading a file on iTunes</u>

<u></u>

Explanation:

  • A feedback mechanism is a loop system wherein the system responds to a perturbation. The response may be in the same direction (as in positive feedback) or in the opposite direction (as in negative feedback).
  • Feedback occurs when outputs of a system are routed back as inputs as part of a chain of cause-and-effect that forms a circuit or loop. The system can then be said to feed back into itself.
  • Evaluation feedback needs to be done “in the moment” to help the person receiving the feedback know where they stand.
  • A feedback control system consists of five basic components: (1) input, (2) process being controlled, (3) output, (4) sensing elements, and (5) controller and actuating devices.
  • Because negative feedback produces stable circuit responses, improves stability and increases the operating bandwidth of a given system, the majority of all control and feedback systems is degenerative reducing the effects of the gain.

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Technician A says independent shops are not affiliated with vehicle manufacturers, but it is easy for technicians who work in th
KatRina [158]

Answer:

b

Explanation:

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8 0
3 years ago
A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen
Lerok [7]

Answer:

Q = 424523.22 kw

Explanation:

\rho =7833 kg/m

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

\alpha = 3.91*10^{-6} m^2/s

T_s = 285 degree celcius

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15

\dot m = 14764.85 kg/s

A_s = \pi DL = \pi 0.4*2 = 2.513 m^2

Q = \dot m C \Delta T = h A_s \Delta T

\dot m C \Delta T = h A_s \Delta T

solving for h

h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}

h = 675.6 kw/m^2K

Q = h A_s\Delta T

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

7 0
2 years ago
Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.
sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

5 0
3 years ago
What’s the answer please help
irina1246 [14]

Answer:

It B

Explanation:

4 0
2 years ago
How to calculate tension.
Evgen [1.6K]

Answer:

Tension can be easily explained in the case of bodies hung from chain, cable, string

Explanation

uniform speed, tension; T = W.

T=m(g±a)

3 0
2 years ago
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