Wednesday, September 22, 2010

Lab 4 Prep #4

14 comments:

Anonymous said...

i cant seem to get the second to last part where it says to use equation 1.6 "The other term involved in determing in the error in the kinetic energy is the error in the sum of the masses and . Equation 1.6 of Lab 1 tells us that (use DeltaM for and Deltam for )"

i wrote down many formulas the one i thought was right goes something like this if anybody can help me out on this
sqrt((DeltaM)^2+(Deltam)^2)
i don't know if its wrong or if my expression is written wrong

Anonymous said...

I have exactly the same problem, I know its correct

Prof. Hobbs said...

The lab prep quiz asks for the uncertainty on the quantity

Delta(M+m)
------------ (1)
(M+m)

but Eqn 1.6 from lab 1 gives the uncertainty on

Delta(M+m).

Don't forget to account for the denominator in (1) above when answering the lab 4 prep questions.

Anonymous said...

right and i have taken that into effect and answered the question before and that answer is not accepted either therefore i began to think maybe it was the wrong expression
this is what i had originally and it was incorrect.
(sqrt(((DeltaM)/M)^2+((Deltam)/m)^2))(M+m)

Anonymous said...

sorry corection (sqrt((DeltaM)^2+(Deltam)^2))*(M+m)

Anonymous said...

then shouldnt we be using equation 1.7 then?

Anonymous said...

(sqrt((DeltaM)^2+(Deltam)^2))/(M+m)

ive been using this formula and it keeps on saying its wrong i dont know what i need to fix in this equation

Prof. Hobbs said...

Instead of (DeltaM)^2, try DeltaM^2. Sometimes these programs are surprisingly fussy about syntax

Anonymous said...

i have tried numerous different answers even those which i knew were wrong and nothing is giving me the correct answer for example i used your advice and put these two answers in and the thing didnt accept it...
(sqrt(DeltaM^2+Deltam^2)(M+m)

(sqrt(DeltaM^2+Deltam^2)/(M+m)

is it possible the program is wrong or our answer is wrong?

Anonymous said...

It took me many tries, but finally this worked... sqrt(DeltaM^2+Deltam^2)/(M+m)

Anonymous said...

thanks a lot, ive been trying all possibilities, but for some reason that worked

Prof. Hobbs said...

The post from Oct 10, 9:23 has an extra unneeded parenthesis at the beginning. (Like energy and momentum, parentheses must be conserved.) The 10:10 PM post does have the correct form.

Anonymous said...

what am i doing wrong? According to 1.8 = abs(n) x Deltad^2/Deltad

Michael Yau said...

According to 1.8 you multiply Deltad/d by the absolute value of n, which is 2 in this case.