• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

Messages
34
Reaction score
4
Points
8
HEY everyone please very important for the practical, the last 2 questions always state 4 sources of errors and improvements, please can someone help me with this


Answers are in the Mark Schemes here: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/

Here are some common answers..
Limitations:
- two results not enough
- measurement of radius/distance/etc. not accurate
- difficult to stop stopwatch at correct moment
Improvements:
- take more readings and plot a graph
- use vernier caliper/travelling microscope
- record using video camera, play in slow motion & use a timer
 
Messages
34
Reaction score
4
Points
8
Someone please answer: Is there a proper angle (from the vertical axis) for the starting position of a pendulum when finding its oscillation's time period? Should the bob be raised at a small angle so it would move slow?
 
Messages
29
Reaction score
8
Points
13
Someone please answer: Is there a proper angle (from the vertical axis) for the starting position of a pendulum when finding its oscillation's time period? Should the bob be raised at a small angle so it would move slow?

no unless mentioned just displace the bob enough so that you can easily track its oscillations and measure its time period
 
Messages
29
Reaction score
8
Points
13
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_21.pdf

4(b)(iii) plsss help!!!

nλ = 2 × 625 is a constant (1250) C1 n = 1 → λ = 1250 outside visible
n = 3 → λ = 417 in visible
n = 4 → λ = 312.5 outside visible
λ=420nm

Really don't understand what the answer says, can someone please explain for me?? Thanks!!!

strange questiono_O okay here goes remember that maxima and minima are formed at alternating points for e.g if n=1 is maxima then n=3,5,7 can only be used for the n values the angle and grating are the same so dsinO is a constant and nλ should be equal to it so just put in the n values in that and find the corresponding wavelenght for this question n=3 gives 417nm so 420nm to 2 s.f hope this helped
 
Messages
382
Reaction score
315
Points
73
can anyone please help with O/N 2008 paper 1 question 11????
The 2kg mass is pulling the 8kg mass with a force of mg = 2 x 9.81 = 19.62N
Hence the force pulling the 8kg mass is 19.62N

But due to friction, the accelerating force would be:
19.62N - 6N
= 13.62N

From Newton's second law of motion;
f = ma
a = f/m
a = 13.62 / (8+2)kg <------ dont forget to add the masses

a = 1.362 m/s^2

Answer = A

Hope that helped :)
 
Top