Ask question

Ask question

All categories

3 years ago

6

During a storm, why do you see the flash of lightning before you hear the roar of thunder? What would it mean if you saw the fla

sh of lightning and heard the roar of thunder at the same time? What would it mean if there were many seconds in between the flash of lightning and the roar of thunder?

1 answer:

Lubov Fominskaja [6]3 years ago

3 0

The speed of light is faster than the speed of sound. So that's why you see a flash of lightning before the roar of thunder. If you heard it at the same time as you saw it, well I will be praying for you. It would literally be exactly where you are. If you hear the thunder a long time after the flash, it means the lightning is far away.

T/5=miles away when T is time from the strike of lightning to hearing thunder

You might be interested in

Find the area of the shaded region

o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

$\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)$

$A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}$

5 0

1 year ago

What values could be substituted for x to make a true number sentence in x + 4 = x + 8

choli [55]

The answer is no solution.

Explanation: x+4 = x+8
-x -x
4 cannot equal 8
Answer is No solution

5 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

2 years ago

Need help, I did a similar problem and got it wrong soo...

Karo-lina-s [1.5K]

Percent change = (new number - old number)/(old number) * 100

percent change = (54 - 48)/(48) * 100 = 6/48 * 100 = 0.125 * 100 = 12.5%

8 0

2 years ago

Read 2 more answers

Use the image below to answer the following question. Find the value of sin x° and cos y°. What relationship do the ratios of si

Licemer1 [7]

Answer:

There are both on the same side of the image "O and P"

If this dose not help please tell me!

7 0

2 years ago

Read 2 more answers