- #1
eNathan
- 352
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[tex].0834 = .5 \sqrt { 1 - v^2 } [/tex]
What is [tex]v^2[/tex]?
What is [tex]v^2[/tex]?
eNathan said:can you explain?
dextercioby said:There's no explanation that could be given by an equation.Explaining means putting words,describing a method...
Daniel.
eNathan said:Ok, I think it is
[tex] v^2 = c^2 \frac {(1-T^2)} {t} [/tex]
mathwonk said:another casualty of the epidemic decline of algebra skills since the introduction of the hand held calculator, and MTV, and the republican controlled congress, and daylight savings time, and fluoridation of the water, and the new deal, and...oops time for my medication.
Since the answer has been provided I'll show you how.eNathan said:Ok, let me ask it in a simpler form.
If [tex]L_t = \sqrt {1 - \frac {v^2} {c^2} } [/tex] then what does [tex]v^2[/tex] equal? BTW is it posible for the explination to be in the form of an anctual equation?
Thx
whozum said:Since the answer has been provided I'll show you how.
[tex] T = t\sqrt{1-\frac{v^2}{c^2}} [/tex]
Divide both sides by t
[tex] \frac{T}{t} = \sqrt{1-\frac{v^2}{c^2}} [/tex]
Square both sides
[tex] \left(\frac{T}{t}\right)^2 = 1-\frac{v^2}{c^2} [/tex]
Subtract one from both sides, then since the right side is negative, multiply both sides by negative one.
[tex] 1-\left(\frac{T}{t}\right)^2 = \frac{v^2}{c^2} [/tex]
Multiply both sides by c^2
[tex] c^2\left(1 - \frac{T^2}{t^2}\right) = v^2 [/tex]
eNathan said:[tex](\frac {T} {t})^2 - 1 = \frac {-v^2} {c^2} [/tex]
Because we divide by [tex]c^2[/tex] on the right of the equation, we can multiply it on the left.
[tex]c^2 (\frac {T} {t})^2 - 1 = -v^2 [/tex]
snoble said:It's the law of distributivity. If you multiply the left hand side by [tex]c^2[/tex] you get
[tex]c^2\cdot ((\frac {T} {t})^2 - 1)= c^2(\frac {T} {t})^2- c^2[/tex]
Think of math like laTex. You don't have to put everything in brackets but if you do it shouldn't change things. When you multiply the left hand side by something you should get the same result whether it is in brackets or it is not. I'm just saying that if you can write tex you can use algebra. Good tex is harder than good algebra.
Steven
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