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julia-pushkina [17]
3 years ago
13

Give two examples of situations or applications where electronic circuits and electric circuits are used together​

Physics
1 answer:
Natalka [10]3 years ago
3 0

The difference between both is that electric circuits makes use of relays, lighting controls, door bells e.t.c, whereas, electronic circuits deal with semiconductors that include transistors and integrated circuits.

Two examples are Smoke Detectors and Electric Toasters.

Now, the examples of two situations where both circuits are used together are;

1) Smoke Detector; In smoke detectors, there is the aspect of power which is the battery used to power the device and that operates with an electric circuit whereas the detector used to detect the smoke and make it beep is based on an electronic circuit.

2) Electric Toaster; The elements of the toaster used to heat materials operates on the principle of electric circuits while the timer used to stop the toasting is based on the principle of electronic circuits.

Read more here; brainly.com/question/24509929

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A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent
Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A  

At all times, angular velocity is \omega = 0\,{\rm{rad/s}  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 0\,{\rm{rad/s}}

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero hence statement A is valid.  

Statement B  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 10\,{\rm{rad/s}}.The final and initial velocities remain the same.  

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement B is valid  

Statement C  

Angular velocity is defined as the change in the angular position with respect to time.  

Angular velocity and angular displacement are related by  

\theta = \omega t

Which can also be modified as:  

{\theta _{\rm{f}}} - {\theta _{\rm{i}}}

Note that the final position is {\theta _{\rm{f}}}and initial position is {\theta _{\rm{i}}}

Modifying the equation to find the angular velocity we obtain  

\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}

When the angular displacement has the same value at all times, the equation becomes  

\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}

The angular velocity becomes zero.  

Angular acceleration and angular velocity are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

The expression above can be rearranged as follows:  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

At all times, the angular velocity is \omega = 0\,{\rm{rad/s}} hence initial and final velocities remain the same  

We obtain  

\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement C is valid.  

Therefore, statements A,B and C are consistent .

4 0
4 years ago
you drive in a straight line at 20.0m/s for 10.0 min, then 30.0m/s for another 10.0 min. Find the average speed.
Juliette [100K]

Answer is 25 m/s.

<em>Explanation;</em>

<em>Step 1 : Convert the units into same unit.</em>

Here speeds are given in m/s while the time is in minutes. So, let's convert minutes into seconds.

1 min = 60 s

Then,

   10 minutes = 10 x 60 s = 600 s

<em>Step 2 :  Find the total distance that you have travelled.</em>

To find that you need to first find distances separately.

We know the formula, s = vt, where s is the distance, v is the speed or velocity and t is the time.

  • First you have travelled 10 minutes (600 s) at 20.0 m/s.

        Then distance, s = 20.0 m/s x 600 s = 12 000 m

  • Secondly you have travelled again 10 minutes at 30.0 m/s

        Then distance = 30.0 m/s x 600 s = 18 000 m

Then, the total distance = 12 000 m + 18 000 m = 30 000 m

<em>Step 3 : Find the total time taken to travel.</em>

Total time  = 600 s + 600 s = 1200 s

<em>Step 4 : Calculate the average speed.</em>

Average speed = total distance / total time

                          = 30 000 m / 1200 s

                          = 25 m/s


   

6 0
3 years ago
Hazards of earthquakes on land and and underwater! (20 points)
elena-s [515]

If the earthquake is on land it would cause total devastation, building will fall ( depending on hos strong it was), people could die but land earthquake is generally more safer than if there was a earthquake underwater. if that happened there would be tsunamis along the coast and everything in its path would be destroyed and your belongings and stuff like cars could be in another town. hope this helps :)

5 0
3 years ago
Read 2 more answers
Are spring oscillations wave transverse or longitudinal
muminat
Longitudinal Waves. .
6 0
4 years ago
The force exerted by the wind on the sails of a sailboat is fsail = 410 n north. the water exerts a force of fkeel = 200 n east.
tangare [24]
The North and East forces are 90° apart, making a right triangle with the resultant as the hypotenuse.
(200)² + (410)² = F²
40,000 + 168,100 = F²
√(208,100) = F
456.18 N = F

F= ma
456.18 N = (300 kg)a
456.18 / 300kg = a
1.52 m/s² = a

The sailboat will be heading North East. To  find the angle of the boats trajectory use inverse tangent function.
tan(Ф) = opposite/adjacent
arctan(opposite/adjacent) = Ф

arctan(410/200) = 64° North East

7 0
3 years ago
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