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3 years ago

What type of modulation is typically used by broadcasting stations to transmit pictures on television screens?

2 answers:

4 0

Analog

Television transmitters use one of two different technologies: analog, in which the picture and sound are transmitted by analog signals modulated onto the radio carrier wave, and digital in which the picture and sound are transmitted by digital signals.

Naddika [18.5K]3 years ago

4 0

Answer:

A. Amplitude

Explanation:

The type of modulation that is typically used by broadcasting stations to transmit pictures on television screens is amplitude. Therefore, the answer to your question is A.

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Find the tension in the two groups that are holding the 2.9 kg object in Pl., One makes an angle of 35.6° with respect to the ve

Stella [2.4K]

Answer:

≈ 20.35 N [newton's of tension]

Explanation:

( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =

( (28.42) ÷ (≈0.813) ) × (≈0.582) =

(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35

4 0

3 years ago

Gauss's law: Group of answer choices can always be used to calculate the electric field. relates the electric field throughout s

kvasek [131]

Answer:

relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss Law states that overall electric flux of a closed surface is equivalent right to charge enclosed which is divided by the permittivity. In other words Gauss Law stress that

net electric flux that pass through an hypothetical closed surface is equivalent to overall electric charge present within that closed surface.

The Gauss law can be expressed mathematically as

ϕ = (Q/ϵ0)

Q = total charge within the surface,

ε0 = the electric constant

5 0

3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago

Find an equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4.

m_a_m_a [10]

An equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4 is x(t)=2t+2,y(t)=t^4.

<h3>What is tangent?</h3>

In calculation, the digression line to a plane bend at a given point is the straight line that "simply contacts" the bend by then. Leibniz characterized it as the line through a couple of boundlessly close focuses on the bend. The chart of digression is intermittent, implying that it rehashes the same thing endlessly. In contrast to sine and cosine in any case, digression has asymptotes isolating every one of its periods. The space of the digression capability is all genuine numbers with the exception of at whatever point cos⁡(θ)=0, where the digression capability is vague. Assuming they stroll in an orderly fashion, they are fundamentally following a digression way for the shape that is made inside the fencing.

Learn more about tangent, refer:

brainly.com/question/12585907

#SPJ4

7 0

2 years ago

Brainliest if correct

poizon [28]

Answer:

Explanation:

hope it helps you make brainliest

3 0

3 years ago

Read 2 more answers