Recent content by JesseK

I'm afraid this is not really relevant to solving the equation for something. However, Fe and Al are the symbols for iron and aluminum. :)

L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] L[0,Fe]α[Fe]*dT - L[0,Al]α[Al]*dT = L[0,Al] - L[0,Fe] dT(L[0,Fe]α[Fe]-L[0,Al]α[Al]) = L[0,Al] - L[0,Fe] dT = L[0,Al] - L[0,Fe] / (L[0,Fe]α[Fe]-L[0,Al]α[Al]) Is this right?

I understand the basic idea that you described in your response, but I'm not entirely sure on how to do that. I can't get all dTs to the left-hand side because L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT --- Subtract by L[0,Fe] L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe]...

Homework Statement L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT How can we solve for dT? I should get those dTs to one side but by dividing they'd just cancel out. L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT -- Subtract both sides by L[0,Fe] L[0,Fe]α[Fe]*dT =...

I have tried to think about it myself as you have probably noticed. Saying that I don't get the idea and asking for help should be completely fine. However, I thank you for giving me this tip.

I don't know how to do it, so could you possibly show me?

cm^2 * kg / m^2 * g m^2 / m = m kg / g = 1000 g => cm * 1000 ?

I've got no idea.

Well 1 km = 1000 meters and 1 hour is 3600 seconds. That gives me the factor.

1 kg = 1000 g kg hg dag g , so 1 0 0 0 30 km/h can be converted into m/s by dividing it by 3,6 , so it'd be 8,333333... m/s However, I don't still get the idea in this specific instance.

How can I solve this problem?

That's right but 1 m^3 can be converted into cm^3, for example. I'd just like to let you know that I still haven't got the idea.

Well, 1 kg/m^3 = 0,001 g/cm^3 --> did you mean it vice versa? If so, then 1 g/cm^3 is 1000 kg/m^3

There are 1000 grams in 1 kg. I still don't get it.

If we count it: h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 333 m It should be 333 meters. How that becomes centimeters? As for your statement, there really isn't much to show. I can't / don't know how to cancel them out but above I explained the idea of canceling out, so I've proved that I know it.