Sarahisme

is that like the 95% confidence interval thing? should i do something such as giving the mean and the 95% confidence level?

anyone? lol, or is thermodynamics not a favourite around here

nevermind, i'll just graph it and take the 3 lines of bestfit. speaking of which, does anyone know a program which will give you the error in your line of best fit?

hey heres the question & heres my answer, but i got stuck so yep any help would be greatly apprectiated! ok here is what i have got so far... [math] \epsilon = 1 - \frac{Q_{out}}{Q_{in}} [/math] [math] Q_{out} \ = \ |Q_{d->a}| \ = \ C_v|T_a - T_d| \ = \ C_v(T_d - T_a) [/math] [math] Q_{in} \ = \ Q_{b->c} = C_p(T_c - T_b) [/math] so [math] \epsilon \ = \ 1 - \frac{C_v(T_d - T_a)}{C_p(T_c - T_b)} [/math] but [math] \frac{C_p}{C_y} \ = \ \gamma [/math] So [math] \epsilon = 1 - \frac{T_d - T_a}{ \gamma (T_c - T_b)} [/math] then using PV = nRT [math] \epsilon = 1 + \frac{1}{ \gamma} \frac{P_aV_a - P_dV_d}{P_cV_c - P_bV_b} [/math] now using the fact that [math] V_a = V_d [/math] and [math] P_c=P_b [/math] [math] \epsilon = 1+ \frac{1}{ \gamma} \frac{V_a(P_a - P_d)}{P_c(V_c-V_b)} [/math] now dividing top and bottom by [math] V_aP_c [/math] [math] \epsilon = 1 + \frac{1}{ \gamma} \frac{\frac{P_a}{P_c} - \frac{P_d}{P_c}}{ \frac{V_c}{V_a} - \frac{V_b}{V_a}} [/math] now [math] PV^{ \gamma } = constant[/math] so [math] P_cV_c^{ \gamma } = P_dV_d^{ \gamma } [/math] then [math] \frac{P_d}{P_c} = \frac{V_c^{ \gamma }}{V_d^{ \gamma }} = ( \frac{V_c}{V_d} )^{ \gamma } [/math] and similarly [math] \frac{P_a}{P_b} = \frac{V_b^{ \gamma }}{V_a^{ \gamma }} = ( \frac{V_b}{V_a} )^{ \gamma } [/math] so [math] \epsilon = 1 + \frac{1}{ \gamma} \frac{( \frac{V_b}{V_a} )^{ \gamma } - ( \frac{V_c}{V_d} )^{ \gamma }}{ \frac{V_c}{V_a} - \frac{V_b}{V_a}} [/math] this is where i get stuck, any suggestions guys n' gals? Sarah

???

ok, but how can the average error be less than all the individual errors? also, what do you guys think of method of getting [math] \pm 6 [/math]

hey i am writing up my prac report i have a question about error caluclation. if i take the average of serveral measurements (instead of drawing a graph), and these measurements are [math] 147 \pm 3 [/math] [math] 146 \pm 3 [/math] [math] 143 \pm 3 [/math] [math] 145 \pm 3 [/math] [math] 146 \pm 2 [/math] i get ther average to be 145 (to 3 sig. figs.) , now how do i calucalte the error? do i use a 95% confiddence interval or somethign else? I have on method i think might work, but it gives a very big error compared to the ones for the orginal measurements: this is the method: [math] ( \delta avg)^{2} = ( \delta m_1)^{2} + ( \delta m_2)^{2} + ( \delta m_3)^{2} + ( \delta m_4)^{2} + ( \delta m_5)^{2} [/math] where [math] m_1 , m_2, ... [/math] are measurement 1, measurement 2,... so [math] (\delta avg) = \sqrt(3^{2} + 3^{2} + 3^{2} + 3^{2} + 2^{2}) = \sqrt(40) = 6 [/math] so avg = [math] 145 \pm 6 [/math] but this error is quite large compared to the others so yeah... any advice would be great! Thanks guys _Sarah

oh ok, yep thanks for that

yeah its cool, i got it in the end, you do it by trying to get lineraly dependent set sort of thing thanks for trying though

hehe :party:

ok heres my attempt at what i think the proof to problem should look like (tell me what you think! ) ok here goes: ----------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------- Define T to be [math] T( \sum_{i=1}^n(a_iv_i)) = \sum_{i=1}^n(a_iw_i) [/math] Since T is defined as transforming the basis vectors of the vector space V, then it is well defined. Now take [math] v, p \in V [/math] and [math] c,d \in \mathbb{R} [/math] (i.e. c,d scalars) So [math] v = \sum_{i=1}^n(a_iv_i) [/math] [math] p = \sum_{i=1}^n(b_iv_i) [/math] [math] (cv + cp) = c(\sum_{i=1}^n(a_iv_i)) + d(\sum_{i=1}^n(b_iv_i)) [/math] [math] = v_i(\sum_{i=1}^n(ca_i + db_i)) [/math] [math] = \sum_{i=1}^n(v_i(ca_i + db_i)) [/math] Then [math] T(cv + dp) = \sum_{i=1}^n(ca_i + db_i)w_i [/math] [math] = c\sum_{i=1}^na_iw_i + d\sum_{i=1}^nb_iw_i [/math] [math] = cT(v) + dT(p) [/math] [math] \therefore [/math] T is linear. [math] \therefore [/math] there is a unique linaer transformation [math] T : V \rightarrow W [/math] such that [math] T(v_i) = w_i [/math] for i = 1,…,n. ----------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------- well that’s what I think it is, however I am not sure this shows that it is unique (or is it because of the fact that is transforms basis vectors that make it unique??, in which case should I say up the top of my proof “Since T is defined as transforming the basis vectors of the vector space V, then it is well defined AND unique.”) well anyway, tell me what you think! Cheers Sarah

So to do this problem do i need to know stuff about difference equations ?? (its a chapter in my linear algebra book) ??

hello?

oh ok i think i see.... a linear transformation maps one vector space into another, right? so if T:E --> W is a linear transformation the range of the transformation is depedent on the vectors in V, and if you have a basis for E, then you effectively have all basis for the range of T, as a linear transformation is defined by T(u+v) = T(u) + T(v) and T(cu) = cT(u) is what i just said correct?....or relevant? Thanks Sarah

"a liner map is uniquely determined by where it sends a basis and specifying the action on the basis uniquely determines the map obviously" is that a thm. of some kind, because i havent seen it in my textbook yet. although i can see that it makes sense.

nevermind, i got it! -Sarah

hmmm....yeah i dunno, i can't find anything in my text book that seems to help me, there is stuff about linear transformations, but sort of dunno i don't see how to apply it to this particular question :S

ok, so now i am thinkning that it has something to do with matrices and linear transformations..... is that the right line of thought?

dammit i wish i knew which damn angle they were talking about!! (i think its the one between the leftmost parrallel line from the virtual image and the axis line????¿¿¿¿¿??????)

umm i think i've got the proof, i'll be back in a few hours, so i'll put it up then

i am not even sure which angle they are talking about on the diagram. :S

ok, cool! , yep i think i got it right this time too (i've checked though it a few times, and it makes sense, and still gives me the same answer each time )

i got i got it! yay!

ok umm so how about i start by well... lol i am not really sure how to start a problem like this, never really done one like it before. how can you show that there is a unique linear transformation? i dunno :S

i think it was this part of my orginal answer that was wrong: we get [math] 1-\frac{v^2}{c^2} = 0.67 [/math] oh and this bit of it too so [math] m = kinetic mass = \frac{E}{c^2} = 8.224\times10^{-28} kg [/math] i guess i just must have been pressing the wrong buttons on my calculator!