Recent content by kimiko333

Great answers! Thank you for the explanations! :) I'm grateful!

Ooooh, okay. Totally makes sense! Great! Thank you! :D Now I just have to figure out, why is it that I can easily solve problems in electrodynamics, thermodynamics, etc, but don't know how to deal with these... :D (I'm a teacher-student, studying to be an English and Physics teacher)

Now it's 12, okay. But if I substitute in, I get 12v, not just simply 12. How do I know, that the answer is in steps?

Oh Jesus. I forgot to multiply v by 10. Aaaahhhh. Told you... I'm tired :D

:biggrin: but why is this so? I'm supposed to get 12. And I get 12 with the second equation

If I substitute 10 into the first equation, I get: ##L=t_1*(\frac {v} {5} +v)## ##L=2v+v=3v##

:O Where?

This is how I'm thinking: I. ##L=t_1*(\frac {v} {5}+v)## II. ##L=t_2*(v- \frac {v} {5})## ##t_1=10 ## ##t_2=15## I. ##L=3## II. ##L=12## Now I'm not so sure what this says to me in a physical sense. Why are there two solutions? I happen to know, that 12 is the correct answer (I had the answer...

Oh, yeah. I forgot to mention that I supposed that t2 is equal to 15 and t1 is equal to 10

Yeah, now I'm thinking about the appropriate equations. But I don't think those are the ones in 19th thread. Because if I substitute in, I get that v=0 :D.

Greeaaat! Finally I found out something :D Now, if I understood correctly. Somehow, I should follow the instructions of the 10th post, right? (and the stress here is on "somehow" :D )

Somehow I found out, that the ratio between v and u is 5 ##\frac {v} {u} =5## Is that correct? :D

Is it ##L=t_2*(u+v)## and ##L=t_1*(v-u)## ?

Great! Okay! Now I feel closer to the solution. Now I can imagine what's going on, and I was amazed by this solving strategy! Still. I have to figure out how to put that into equations.

Oh, okay. That makes sense!