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Naily [24]
3 years ago
8

Can someone write a 200 speech that explains the difference between analog and digital waves and which one is better?

Physics
1 answer:
svetoff [14.1K]3 years ago
3 0

Answer:

Analog and digital signals are the types of signals carrying information. The major difference between both signals is that the analog signals have continuous electrical signals, while digital signals have non-continuous electrical signals.

Explanation:

Analog Signals   were used in many systems to produce signals to carry information.  These signals are continuous in both values and time. The use of analog signals has been declined with the arrival of digital signals. In short, to understand the analog signals –  all signals that are natural or come naturally are analog signals.  Digital Signals , unlike analog signals, digital signals are not continuous, but signals are discrete in value and time. These signals are represented by binary numbers and consist of different voltage values.  Their differences are as follows:

Analog Signals                                      Digital Signals

Continuous signals                               Discrete signals

Represented by sine waves               Represented by square waves

The Human voice, natural sound, analog electronic devices are few examples   Computers, optical drives, and other electronic devices

A Continuous range of values            Discontinuous values

Records sound waves as they are     Converts into a binary waveform.

Only be used in analog devices.        Suited for digital electronics like computers, mobiles, and more.

I think a Digital signal is better.

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lutik1710 [3]

You need to observe the car at two different times.

-- The first time: 
You write down the car's speed, and the direction it's pointing.

-- The second time:
You write down the car's speed and the direction it's pointing, again.

You take the data back to your lab to analyze it.

-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.

-- You compare the first and second direction.  If those are different,
even if the speeds are the same, then the car had acceleration during
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(Remember, "acceleration" doesn't mean "speeding up". 
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8 0
3 years ago
Read 2 more answers
In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c
sammy [17]

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car        v = 278 m/sec

Constant acceleration a=5.56m/sec^2

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion v^2=u^2+2as

Putting all values in equation

278^2=0^2+2\times 5.56\times s

s = 6950 m

So distance traveled during this acceleration will be 6950 m

3 0
3 years ago
An object has a mass of 5.20g. The mass of the object in kg is​
vampirchik [111]

Answer:

0.0520kg

Explanation:

divide by 1000

8 0
2 years ago
A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

3 0
3 years ago
Which of the following people challenged the geocentric model of the solar system?
Neporo4naja [7]
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6 0
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