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cupoosta [38]
3 years ago
12

Open the site in your browser. (If any pop-up windows appear, think about what they indicate about the web site before you close

them.) Skim the article. Think about whether you trust the information presented or whether you have doubts about some of it. On a scale of 0 to 6, where 6 is the most trustworthy, how would you rate this site? (Note that all responses will be marked as "correct" at this point.)
Computers and Technology
1 answer:
kati45 [8]3 years ago
5 0

Answer:

The answer to the given question is 5-6 (trustworthy).

Explanation:

When visits any web site there are several information is given, We click on it shows some pop-up box. it is smaller than the whole windows. for example, If we want to download any free software. So we go on a software site and choose the version for software and click for download. when we click on download there is a pop-up window open to provide the details of the software and has a one-button when we click on the button it downloads the software. If we cancel the pop-up window by click on the cancel /close (X) button it will close the window.

So the pop-up windows are very trustworthy.

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#include<stdio.h>

#include<stdlib.h>

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                   break;

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                   fscanf(fp,"%c",&ch);

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3 0
3 years ago
Read 2 more answers
Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
Ede4ka [16]

Answer:

t= 8.7*10⁻⁴ sec.

Explanation:

If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

v = \frac{(xf-xo)}{(t-to)}

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

t =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

⇒ t = 8.7*10⁻⁴ sec.

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