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2 years ago

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

1 answer:

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

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3 years ago

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3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,