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survivorboi · Jul 21, 2009, 02:25 PM

Let's pick an equation for an example, let's pick Einsteins E=mC^2.
When I see an equation like that, I don't know how to use it or solve it. I'm confused on whether I should substitute the letters for the correct numbers or what? Like in the equation above, e=energy m=mass c=speed of light

So, if I have something with the mass of 2 grams and speed of light is 186,000 miles/hr, do I just substitute them?

E=(2*186,000)^2?

Then I get E?

It's the same with other equations, I'm confused on how to use it... Like Newton's equation:

F=(G*m1*m2)/r^2

Do I just substitute letters for numbers?

galactus · Jul 21, 2009, 03:56 PM

You can use Einstein's famous equation and solve like any other. Do you know what this equation means?

Anyway, here is an example problem.

"The power output of the Sun is huge. $3.8\times 10^{26}$ watts.

How much rest mass is converted to kinetic energy in the Sun every second"?

Using $E=mc^{2}$

$c=3.0\times 10^{8} \;\ \frac{m}{s}$ is the speed of light in meters per

second.

$m=\frac{E}{c^{2}}=\frac{3.8\times 10^{26} \;\ J}{(3.00\times 10^{8} \;\ \frac{m}{s})^{2}}=4.2\times 10^{9} \;\ kg$

You can use 186,000 miles per second. Just make sure you use consistent units

with the mass or energy as well. Often, you will see the speed of light in meters per

second instead of miles per second

In your example, you are using grams and miles per second... inconsistent.

Try this: $(.002 \;\ kg)(3.0\times 10^{8} \;\ \frac{m}{s})^{2}=1.8\times 10^{14} \;\ J$

This is just a very basic example using this wonderful equation.

survivorboi · Jul 21, 2009, 09:32 PM

Originally Posted by galactus

You can use Einstein's famous equation and solve like any other. Do you know what this equation means?.

Anyway, here is an example problem.

"The power output of the Sun is huge. $3.8\times 10^{26}$ watts.

How much rest mass is converted to kinetic energy in the Sun every second"?.

Using $E=mc^{2}$

$c=3.0\times 10^{8} \;\ \frac{m}{s}$ is the speed of light in meters per

second.

$m=\frac{E}{c^{2}}=\frac{3.8\times 10^{26} \;\ J}{(3.00\times 10^{8} \;\ \frac{m}{s})^{2}}=4.2\times 10^{9} \;\ kg$

You can use 186,000 miles per second. Just make sure you use consistent units

with the mass or energy as well. Often, you will see the speed of light in meters per

second instead of miles per second

In your example, you are using grams and miles per second....inconsistent.

Try this: $(.002 \;\ kg)(3.0\times 10^{8} \;\ \frac{m}{s})^{2}=1.8\times 10^{14} \;\ J$

This is just a very basic example using this wonderful equation.

Let me try to understand this. This is what I thought I have to do: Since we are trying to figure out E=energy, we don't know it yet. M=mass, so let's say if a ball is 4 unit mass, and c^2=speed of light square. Saying all that, shouldn't the equation be this:

E=4(186,000^2)?

After figuring the right side out, you get E? And that's the answer.

Because the way you explained it to me was a lot more complex. And, what do you mean by inconsistent?

Unknown008 · Jul 22, 2009, 01:28 AM

Ok, I'll answer since I'm here.

You did:

E=(2*186,000)^2?

That's not it.

Firstly, all the units have to be 'on the same level'. If you're using SI units, use them everywhere. For example, if you're using joules, use kilograms and metres per second. If you're using millijoules, use grams and millimetres per second.

Secondly, it is $E = mc^2$ or for better understanding to you, $E= (m)(c^2)$

In algebra, the variables are 'detacted' from each other concerning powers.

Say you have $2x^2$ . That doesn't mean replacing x by 2 will give $[2(2)]^2 = 16$ but it will give $2(2^2)=8$

Is that what you were asking?

survivorboi · Jul 22, 2009, 03:18 PM

Originally Posted by Unknown008

Ok, I'll answer since I'm here.

You did:

That's not it.

Firstly, all the units have to be 'on the same level'. If you're using SI units, use them everywhere. For example, if you're using joules, use kilograms and metres per second. If you're using millijoules, use grams and millimetres per second.

Secondly, it is $E = mc^2$ or for better understanding to you, $E= (m)(c^2)$

In algebra, the variables are 'detacted' from each other concerning powers.

Say you have $2x^2$ . That doesn't mean replacing x by 2 will give $[2(2)]^2 = 16$ but it will give $2(2^2)=8$

Is that what you were asking?

So, the equation is not:

E= (MC)^2?

But:

E= (m) (c^2)?

All right, here i go

Since I used gram which is metric, I can't use miles for the speed of light, so I shall use kilometers.

Speed o light: 299792.458 km/s

So now my equation is :

E= 4(299792.458^2)

RIGHT?

So E= 359502071494.727056

What is THAT suppose to mean?

Unknown008 · Jul 23, 2009, 03:27 AM

Wait wait wait...

So, if I have something with the mass of 2 grams and speed of light is 186,000 miles/hr, do I just substitute them?

The SI unit for mass is kilogram. So, a gram is 1/1000 of a kilogram.
The Si unit for distance is metre. So, a kilometre is 1000 times a metre.

These are not 'in accordance'! You should have taken millimetre if you use grams. Or you use kilograms with metres and tonnes with kilometres.

So, the equation is not:

E= (MC)^2?

But:

E= (m) (c^2)?

Yup, the brackets are not shown but they are there.

E= 4(299792.458^2)

Where did you obtain 4? You again did the square of '2' or you took another mass of 4?

Ok, now the answer was already given by galactus,

Originally Posted by galactus

In your example, you are using grams and miles per second... inconsistent.

Try this: $(0.002 \, kg)(3.0 \times 10^8 m/s)^2=1.8 \times 10^{14} \, J$

What is THAT suppose to mean?

That means that such a mass (if all it's energy were to be removed) would give that amount of energy.

2260 J of energy is needed for 1 kg of water to completely boil at 100 Celsius.
Can you estimate the volume of water you can boil with so many joules of energy?

survivorboi · Jul 23, 2009, 02:33 PM

Originally Posted by Unknown008

Wait wait wait...

The SI unit for mass is kilogram. So, a gram is 1/1000 of a kilogram.
The Si unit for distance is metre. So, a kilometre is 1000 times a metre.

These are not 'in accordance'! You should have taken millimetre if you use grams. Or you use kilograms with metres and tonnes with kilometres.

Yup, the brackets are not shown but they are there.

Where did you obtain 4? You again did the square of '2' or you took another mass of 4?

Ok, now the answer was already given by galactus,

That means that such a mass (if all it's energy were to be removed) would give that amount of energy.

2260 J of energy is needed for 1 kg of water to completely boil at 100 celcius.
Can you estimate the volume of water you can boil with so many joules of energy?

Umm... 4 was the mass of the object in grams.

So, can you please explain what you meant when u said things like when I use grams, I have to use kilometres, etc?

Are you trying to say that if I use grams for mass, then I have to use what for the speed of light? And other scenarios...

I didn't get what you meant by using the "same" unit? Grams and kilometers are metric system ,which is the same type isn't it?

Unknown008 · Jul 25, 2009, 01:07 AM

Umm... 4 was the mass of the object in grams.

Ok, so you took another mass.

So, can you please explain what you meant when u said things like when I use grams, I have to use kilometres, etc?

Are you trying to say that if I use grams for mass, then I have to use what for the speed of light? And other scenarios...

I didn't get what you meant by using the "same" unit? Grams and kilometers are metric system ,which is the same type isn't it?

The SI units are the kilogram for mass, the metre for distance, the second for time, the ampere for current, the kelvin for temperature, the mole for the amount of substance and the candella for light intensity.

When you use kilograms, you'll have to use metre, second, etc...
If you use grams, you'll have to use millimetre, millisecond, milliampere, etc
If you use tonnes (1000 kilograms) you'll have to use kilometre (1000 metres), etc.

So, in the case of $E = mc^2$ , it's better convert everything to the SI units, that is energy in joules, mass in kilograms, speed in m/s.

If you use grams, you'll have to use millijoules, millimetres per second, mm/s

If you use tonnes, you'll have to use kilojoules, kilometres per second, km/s.

Is it clearer now?

It's the same for other formulae;

Distance (m)= Speed (m/s) x Time (s)

You cannot have distance in metres and speed in kilometres per hour and time in miunutes. You'll have to convert kilometres per hour into metres per second and minutes to seconds before proceeding.