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Ket [755]
2 years ago
6

What are the advantages of using digital signals over analog signals?

Physics
1 answer:
Snezhnost [94]2 years ago
4 0

Digital signals are advantageous over analog signals because they are easily reproducible and convey information with little interference.

<h3>What are analog and digital signal?</h3>

Analogs signals are signal which are continuous whereas digital signals are signal which are not continuous but are separated by time.

Use of digital signals are most common because they have certain advantages over analog signals.

Some advantages of digital signals over analog signals include:

  • Digital signals can convey information with less noise and interference.
  • Digital circuits can be easily reproduced

Therefore, Digital signals are advantageous over analog signals because they are easily reproducible and convey information with little interference.

Learn more about digital and analog signals at: brainly.com/question/14729899

#SPJ1

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Serhud [2]
<span>According to Newton's first law of motion:

-- objects at rest will remain at rest unless acted upon by an outside force

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3 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
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umka2103 [35]

Answer:

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Explanation:

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5 0
3 years ago
Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,
yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

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3 years ago
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<span>the most important personality trait the coach should demonstrate to achieve a goal is </span>persistence 
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