The formula to determine energy is E=1/2mv^2 . What is the formula solved for v?
\(E=\Large \frac{1}{2mv^2}\) You are solving for v in this equation, but this equation is solved for E. Do you have any ideas on where to begin?
No idea honestly. I just know the answer..
multiply times the inverse any fraction in your case 1/2 -> inverse = 2 divide over other variables , keep what you want take the square root neglect the negative root.
This is kinetic energy \[KE = \frac{ 1 }{ 2 }mv^2\]
\[\color{red}{2} \times KE =(\frac{ 1 }{ 2 }mv^2) \times \color{red}{2} \implies 2 KE = mv^2\] see how that works so far?
Yes. Multiply by the inversion to eliminate the 1/2
How can we get rid of the m on the right?
ummm... I would think Subtraction property?
Why, notice it's being multiplied by the v^2
What if we did this \[\color{red}{\frac{ 1 }{ m }} \times 2KE = (mv^2) \times \color{red}{\frac{ 1 }{ m }}\]
\[\implies \frac{ 2 KE }{ m } = v^2\] so we have this, then what?
then would need to eliminate the Coefficient and the exponent.??
We need to get rid of the exponent try this, \[x^2 = something \implies x = \sqrt{something}\]
hmmm.. so wait.. Relating to the \[v=\pm \sqrt{\frac{ 2KE }{ m }}\].. how do you eliminate the K?
The KE just means kinetic energy, I replaced the E with KE to show the difference as we usually use E for total energy you can leave it as E however :)
ohhhh ok. so the answer is indefinitely \[v=\pm \sqrt{\frac{ 2E }{ m }}\]
Yes, that's fine :)
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