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Is this actually a physics or math project or just a question asking for you to solve for the variable "t"?
you can use basic algebra to solve for t by isolating it, but it will be an equation using L. But that might be all you need to do... sort of depends on what the purpose of this question is.
just have to solve for t.. i think :/ sorry my computer is lagging!

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Other answers:

\[L = \frac{ 980t^{2} }{ 4\pi^{2} }\]
yeah, that's it :)
  • phi
here is an example http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/v/solving-for-a-variable The first step is multiply both sides by 4\(\pi\). Can you do that?
  • phi
The first step is multiply both sides by 4\(\pi^2\). Can you do that?
hmm, not really.. :///
  • phi
write \(4 \pi^2\) times on both sides of the equation
okay
like i dont understand it because it has the pi? :/ i suck at math
  • phi
pi is just a number if you multiply both sides by 4 pi^2 you get \[ 4 \pi^2 L = 4 \pi^2 \cdot \frac{980 t^2}{4 \pi^2}\]
  • phi
now use a simple rule: if you see something divided by itself , it becomes 1. Does that ring a bell? It makes the right side "simplify"
no it doesnt :(
ahhh, this is confusing to me
  • phi
Here are some examples \[ 3\cdot \frac{4}{3} = \frac{3\cdot 4}{3}= \frac{\cancel{3}\cdot 4}{\cancel{3}}=4\]
  • phi
do you see anything in your problem that "cancels"?
yeah, i get that but it has the pi in the one im doing? :/
so the pi would cancel out?:S:/
  • phi
\[ 4 \pi^2 L = \cancel{4 \pi^2} \cdot \frac{980 t^2}{\cancel{4 \pi^2}} \]
yeaah, thats what i thought :))
  • phi
it is a good rule to remember. now multiply both sides by \(\frac{1}{980} \)
one over what? i can't read that:((
  • phi
1/980
  • phi
first write it down: \[ 4 \pi^2 L \cdot \frac{1}{980}= 980t^2 \cdot \frac{1}{980} \]
k 1 sec
okays (:
  • phi
the 980's cancel on the right side, right? (that is why we did this) we get \[ t^2 = \frac{4\pi^2 L}{980} \] Do you see how we did that?
ohhh, yeah i did
  • phi
the 4 pi^2 L is on the top, and when we multiply fractions, it is top times top and bottom times bottom
  • phi
\[ \frac{4 \pi^2 L}{1} \cdot \frac{1}{980} = \frac{4 \pi^2 L\cdot 1}{1 \cdot 980} = \frac{4 \pi^2 L}{980}\]
  • phi
I would simplify by dividing the top by 4, and the bottom by 4. Can you do that?
all of the top?:S i dont know how to simplify because the pi is there? do i use 3.14?:/
  • phi
leave pi alone. if you divide the top by 4 it is doing this \[ \frac{\cancel{4}\pi^2L}{\cancel{4}}\]
  • phi
now divide the bottom by 4: 980/4 is 245 we get \[ t^2= \frac{\pi^2 L}{245} \]
  • phi
do you know how to "solve" for t?
multiply both sides by 245?:S
im not sureee :(
  • phi
if you multiply both sides by 245 you would get a 245t^2 (making it more "complicated" ) Have you heard of a square root?
yeaah i have
  • phi
square roots are how you "undo" squares: \[ \sqrt{x^2}= x\]
  • phi
I would take the square root of both sides
can you draw it out for me so i can see what you mean? it makes more sense when you do that:D
  • phi
I've noticed it's easier to do work when someone else is doing it, too. try, it won't hurt
it really does help when someone else shows you, but ill draw it on here to see if im right :) im prob not thou ...haha
  • phi
Just draw a big square root over each side
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  • phi
yes, by the t^2 is inside the root sign. and so is the 245. now use the idea that sqrt(x^2) is x so sqrt(t^2) is ?
245? haha.. im not sure :/ or 15? :S
  • phi
no, not a number. square root of x*x is x square root of t*t is t (I guess you have to memorize that) we get \[ t= \sqrt\frac{ \pi^2 L}{245} \]
  • phi
now, you can "simplify" (just to show off how much math you know) use the rule that sqrt(a * b) is the same as sqrt(a) * sqrt(b) so the top is sqrt(pi^2) * sqrt(L) you can simplify sqrt(pi^2) , right?
yes?.....
  • phi
use the same rule as for sqrt(x^2) (or sqrt(t^2) for that matter)
i dont understand when you say sqrt)x^2) really..
  • phi
we start with \[ t= \sqrt\frac{ \pi^2 L}{245} \] we can "break up" the square root by separating it: \[ t = \frac{ \sqrt{\pi^2}\sqrt{L}}{\sqrt{245}} \] the rule "sqrt(x^2) = x means when you see "something" to the 2nd power inside a square root , you can replace it with just "something" There is a good reason to learn this, but first what is \[ \sqrt{\pi^2} \] ?
  • phi
or, another way to think of it, sqrt undoes the square
ooh, that kinda makes more sense. what do you mean what is it?
so it would be pi^2?
or sqaure root pie? :S
or neither? hahaa
  • phi
sqrt(pi^2) is not pi^2 (that would mean sqrt is doing nothing) sqrt(pi^2) is not sqrt(pi) (that would mean pi^2 is the same as pi) sqrt(pi^2) "makes the square go away"
  • phi
if \[ \sqrt{x^2}= x\] and \[ \sqrt{t^2}=t\] what is \[ \sqrt{\pi^2}= ?\]
pi..lol
  • phi
yes, so now you have \[ t = \frac{ \sqrt{\pi^2}\sqrt{L}}{\sqrt{245}} \] \[ t = \frac{ \pi\sqrt{L}}{\sqrt{245}} \] finally we can rewrite sqrt(245) as sqrt(5*7*7) or \[ \sqrt{7^2}\cdot \sqrt{5} \] do I dare ask what is sqrt(7*7)?
hahah!
49
  • phi
7*7 = 49 but sqrt(7*7) is 7 (or sqrt(49) is 7) 7*7 is 7^2 and the square root undoes the square.
yeaah
  • phi
we now have \[ t = \frac{ \pi\sqrt{L}}{\sqrt{245}} \] \[ t = \frac{ \pi\sqrt{L}}{7\sqrt{5}}
  • phi
\[t = \frac{ \pi\sqrt{L}}{7\sqrt{5}} \]
  • phi
The last step. People (why?) do not like sqrt in the denominator, so we multiply top and bottom by sqrt(5) \[ t = \frac{ \pi\sqrt{L}}{7\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}= \frac{ \pi\sqrt{5L}}{35}\]
  • phi
that is your answer
wow, that is a loooot of work!! but thank you so much.. when i re-read over it tomorrow, i'll prob be awesome at it ;) haha....ill prob get it! :D are you a teacher?
  • phi
The reason you learn this is to 1. See if you *can* learn it. 2. If you can learn it, it changes how you think (in a good way)
  • phi
I am an engineer for a living..
thats awesome :D

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