Solved physics Paper 5??

[xhizors]

some uncertainties help
note
# equals change

1.when finding error for logs
log(worst value) - log(best value)

2. when square
#x^2/x^2 = 2(#x/x)

3. when in formula
say-> Y= x/z
so error=>
#y/y= #x/x + #z/z

4. when finding gradient
diff of best gradient and worst gradient

5. when y-intercept errors
find y-intercept with bst gradient than with worst gradient find their diff that is your uncertainity

mostly same is the case with the last one

hope this helps

 

[studen12345]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_53.pdf
how to find uncertanity in v^2 in Q2 plzz answer tahnks

 

[Ashrith]

Can someone tell me how to do the table for Q2 of 9702/53/O/N/12?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_53.pdf

 

[fb.junks]

Instead of a cross, is just a dot allowed?

A dot is allowed but u have to circle it too!!!!

 

[fb.junks]

Can someone please help me with Q.2 (e)? It is really important and I would appreciate help immensely!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_53.pdf

V=V0e^(-t/CR)
.10V0=Voe^(-t/CR) (V0 cancelled on both sides of the equation)
.1=e^(-t/CR)
ln(.10)=-t/CR
t=15
Use the value of C and T to find R.
Error of R=(error of C/C)xR

 

[Soldier313]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_53.pdf
how to find uncertanity in v^2 in Q2 plzz answer tahnks

Can someone tell me how to do the table for Q2 of 9702/53/O/N/12?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_53.pdf

v = s/t where s=distance, t=time
v^2 = (s/t)^2
so v^2 = (0.05/t)^2
then substitute values of time from each row into the eqn
eg for row 1, v^2 = (0.05/0.046)^2
so v^2 = 1.18


then uncertainty is (max value of v^2 - min value of v^2) / 2
so max value of v^2 = (0.051/minimum time)^2
min value of v^2 = (0.049/ max time ) ^2
subtract these two values and then divide by 2 to obtain uncertainty


Hope that helped

 

[fb.junks]

A small dot will work

U have to circle it too

 

[lavanyamane]

V=V0e^(-t/CR)
.10V0=Voe^(-t/CR) (V0 cancelled on both sides of the equation)
.1=e^(-t/CR)
ln(.10)=-t/CR
t=15
Use the value of C and T to find R.
Error of R=(error of C/C)xR

That was brilliant! I cannot thank you enough. All the best for your exam tomorrow

 

[chocolatelover]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s08_qp_5.pdf
Range of ammeter in question 1.
They have suggested in additional details.
And they have given the resistivity. But idk how to use ρ and find the range of ammeter

 

[lavanyamane]

Could anyone please tell me what the likely ranges of an ammeter and a voltmeter are? Thankay xx

 

[Nibz]

You know, I did this. I did the same exact thing that you did. But then, why does the Mark Scheme say something else?

Because in the mark-scheme, the examiner has only given the values for the 'mass ratio row' and 'v2 row', not the 'v-row'.

 

[aehn laseem]

can somebody explain to me wat is a compact light source,,plz...I'd really appreciate t

 

[FRENZYAMU]

can somebody explain to me wat is a compact light source,,plz...I'd really appreciate t

The Compact Light Source (CLS) is a breakthrough technology that addresses the increasing demand for access to high quality X-rays by offering the possibility of a synchrotron beamline for home laboratory applications

googled it

where did u read the :compact light source" :O

 

[aehn laseem]

The Compact Light Source (CLS) is a breakthrough technology that addresses the increasing demand for access to high quality X-rays by offering the possibility of a synchrotron beamline for home laboratory applications

googled it

where did u read the :compact light source" :O

i came across it somwhere..but thanks for the effort..

 

[Nibz]

Could anyone please tell me what the likely ranges of an ammeter and a voltmeter are? Thankay xx

Ammeter range should be about 0-5 A.
And voltmeter's range should be about 0-10 V.

Double check this, though.

 

[Iadmireblue]

v = s/t where s=distance, t=time
v^2 = (s/t)^2
so v^2 = (0.05/t)^2
then substitute values of time from each row into the eqn
eg for row 1, v^2 = (0.05/0.046)^2
so v^2 = 1.18


then uncertainty is (max value of v^2 - min value of v^2) / 2
so max value of v^2 = (0.051/minimum time)^2
min value of v^2 = (0.049/ max time ) ^2
subtract these two values and then divide by 2 to obtain uncertainty


Hope that helped

Is it correct to find find the percentage uncertainity in both and multiply it by two,and use this answer to find absolute error?
and why divide by two?
Thanks

 

[chocolatelover]

Hey my error in gradient is 200 :S
I guess that's the reason why we got such different values for the % error

Your approach is correct, but i guess you should recheck your error in gradient, i checked mine, seems correct :/

Okay I checked it twice mine is not going under 700. :O
I used (33.3, 2.06) and (150, 1.33) to take out the gradient.
Proof.

 

[chocolatelover]

Do we place the Hall probe inside a coil?

 

[blueberryyums]

Must we circle the dot? I never circle the dot. Most graphs I see on here too do not circle the dots. What to do?!

 

[Soldier313]

Okay I checked it twice mine is not going under 700. :O
I used (33.3, 2.06) and (150, 1.33) to take out the gradient.
Proof.

Hey, i think you forgot to include the exponent -6 in the denominator of your gradient calculation