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3 years ago

What questions should be considered when evaluating the reliability of a website?

2 answers:

6 0

How accurate a website is?
Who is the author?
Is the website info being bias?
When was the website written and is it recent?
Does the web have and mistakes?
Is there a way to contact?
Is there bibliography given?
Is there information given on the author?
Is the author a reliable?

Alex787 [66]3 years ago

5 0

Explanation:

Is the site from an established institution or reputed organization?

Who is providing the information?

What is the author's credential?

Is the site bias?

Is the site trying to influence you?

Is the author's contact information provided?

When the site was last updated?

When the site was first created?

What is the major domain of the web address?

You might be interested in

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

3 years ago

What is the energy (in J) stored in the 14.0 µF capacitor of a heart defibrillator charged to 6.90 ✕ 103 V?

Eduardwww [97]

Answer:

The energy stored in the capacitor is 333.3 J

Explanation:

Energy stored in a capacitor = CV²/2

C = capacitance = 14 × 10⁻⁶F

V = 6.90 × 10³V

Energy = (14 × 10⁻⁶)(6.90 × 10³)²/2 = 333.27 = 333.3 J

Hope this Helps!!!

6 0

4 years ago

A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates

tensa zangetsu [6.8K]

Answer:

Electric field at a distance of 1.45 cm will be $172.41\times 10^4N/C$

Explanation:

We have given the distance d = 1.45 cm = 0.0145 m

And the potential difference $V=2.5\times 10^4volt$

There is a relation between potential difference and electric field

Electric field at a distance d due to a potential difference is given by

$E=\frac{V}{d}$ , here E is electric field, V is potential difference and d is distance

So $E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C$

8 0

4 years ago

What is the average velocity of a dragster that starts from rest at the starting line and finishes a race at 2600 m/s in 12 seco

xxMikexx [17]

Using the equation v(avg)=distance/time
and the equation v=v(original)+a(t)
solve for acceleration
2600=0+a(12)
a=216.66666 m/s^2

Then, you use the equation
v^2=v(original)+2a*(change in x)
2600^2=2(216.666666)*change in x
6760000/2/216.666666 = 15600 meters which is the length of the race

Then using v(avg)=x/t
15600/12= 1300 m/s

7 0

3 years ago

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t

Salsk061 [2.6K]

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{\frac{-bt}{2m}}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{\frac{-bt}{2m}}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

8 0

3 years ago