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1 year 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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What force is affected when the distance between two objects remains the same and the mass of each object is doubled?.

rusak2 [61]

Answer:

Gravitational force between masses there is Gravitational force .

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2 years ago

A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti

Kitty [74]

Answer:

$2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4$

$4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2$

Explanation:

Given values

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

<u>The law of conservation of energy</u>:

$\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }$

$\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }$

$\frac{k \Delta l^{2}}{2}$ Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

$\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}$

Where, Δl is initial spring deformation

$\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}$

$\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}$

$\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}$

$v=\sqrt{\frac{400}{2} \times(0.3)^{2}}$

$\mathrm{v}=\sqrt{200 \times 0.09}$

<u>The law of conservation of energy</u>:

$\frac{m v^{2}}{2}+m g h=\text { constant }$

Where h is height

$\frac{m v^{2}}{2}+0=0+m g h$

$\frac{m v^{2}}{2}=m g h$

Cancel mass "m" each side

$\mathrm{h}=\frac{v^{2}}{2 g}$

Distance along incline equals

$\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}$

$\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}$

$\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}$

$\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}$