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http://keisan.casio.com/exec/system/1224060366

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- Thread starter Noah Englehart
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- #1

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http://keisan.casio.com/exec/system/1224060366

- #2

jtbell

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What specifically is confusing you about it?

- #3

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Well.. I understand the square root part, because that is in all the formulas, but i don't exactly understand what the m0 means, and i have no clue as to how to get the E0. Oh, And sorry, I forgot to post the website. Posting now.What specifically is confusing you about it?

- #4

Mister T

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##m_o## is the mass. ##E_o## is the rest energy. ##E_o=m_oc^2##.

- #5

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I know all that, The problem is,I do the square root part... And then im not sure how to divide that. Because, I multiply the mass with speed of light, divide the result with the square root part, And then run it through the calculator to find its inconsistent with my results.##m_o## is the mass. ##E_o## is the rest energy. ##E_o=m_oc^2##.

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- #6

jtbell

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What square root?The problem is,I do the square root part...

- #7

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The square root at the bottom part of the fraction.What square root?

- #8

jtbell

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Argh. I finally clicked your link. Sorry about that. Now I think I see what you're asking about.

How about showing us exactly what numbers you're trying to plug in, and the intermediate steps of your arithmetic? Then someone can probably tell you where you're going wrong.

- #9

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Well, I know that when i do the square root, its right, because the calculator confirms. However, I then tried to divide m0 x c2, which worked. So then i declared that must be the value of E0, But then i get confused because i realize that the E0 is a fraction and that messes up my thought processes.

Argh. I finally clicked your link. Sorry about that. Now I think I see what you're asking about.

How about showing us exactly what numbers you're trying to plug in, and the intermediate steps of your arithmetic? Then someone can probably tell you where you're going wrong.

- #10

HallsofIvy

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[tex]E=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

If you take, say, [itex]m_0= 1 kg[/itex] and [itex]v= c/2= 299792.5/2=149896.25 km/s.[/itex] then that formula says that

[tex]E= \frac{c^2}{\sqrt{1- \frac{1}{4}}}= \frac{c^2}{\frac{\sqrt{3}}{2}}= \frac{2c^2}{\sqrt{3}}[/tex] which would be 103779337954.2 Joules.

That is, much to my surprise, what I get when I enter the values into that website.

- #11

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[itex] E = \frac{E_o}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{m_oc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] is the energy of m

- #12

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I don't know if anyone will be able to help until you post your calculations step by step.Well.. I understand the square root part, because that is in all the formulas, but i don't exactly understand what the m0 means, and i have no clue as to how to get the E0. Oh, And sorry, I forgot to post the website. Posting now.

Regardless, E

- #13

Khashishi

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$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

so

$$E = \gamma m c^2$$

It just means that an object moving at a faster velocity has more energy than an object at rest. How much more? Well that's what the formula tells you. The kinetic energy T is equal to the total energy minus the rest energy.

$$T = E - E_0 = (\gamma-1) m c^2$$

Note that for small ##v##, ##\gamma-1 \approx \frac{1}{2} \frac{v^2}{c^2}## (why don't you verify this with some trial numbers, or Taylor expansion if you know how?) so ##T \approx \frac{1}{2} m v^2##

- #14

Khashishi

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$$E^2 = m^2 c^4 + p^2 c^2$$

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