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ALL CIE PHYSICS DOUBTS HERE !!!

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$!$RatJumper$!$:
W05 P2
Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
 
Thank you

birchy33:
yup spot on. Paper 2.

astarmathsandphysics:
when I gwt bup shortly

astarmathsandphysics:
Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
3c)36 degrees below the horizontal, at right angles to the upper string
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
5vii)amplitude of A=sqrt(1) and amplitude of B=sqrt(4/9)=2/3
1-2/3=1/3 then 1/3^2=1/9 . Subtract since there are 180 degrees out of phase

$!$RatJumper$!$:

--- Quote from: astarmathsandphysics on November 08, 2010, 09:22:18 am ---Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain

3c)36 degrees below the horizontal, at right angles to the upper string
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
5vii)amplitude of A=sqrt(1) and amplitude of B=sqrt(4/9)=2/3
1-2/3=1/3 then 1/3^2=1/9 . Subtract since there are 180 degrees out of phase

--- End quote ---

i really dont understand 3 and 5.. please could you elaborate :)
and fir 4, the mark scheme says horizontal.. doesnt that mean 90 degrees? so why is it 36?

thankx a lot

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