Oh great thanks!
main marks are for main logic applied to the question !!! yes it is possible
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main marks are for main logic applied to the question !!! yes it is possible
yes .. i got a question !!!I guess it will be 40 maybe 39 but probably 40..
it was Q 2 !!! in which V and H was to be fooundWhat answer did you guys get for finding V in question 4?? please tell me
V was 0.966not that one..the question on particle P something like that ( the question about plane A,b..)
in Q2... what was V and H ??Yeah
in q2 v was 40 and h was 80 !!2)V was 30 something
H was 60 Something
3)t was 60 and 40 or something and S was 4hundred something
4)V was something
5) A was 2/3 T was 1.07 and length of string was 1.9something
6) mew was 0.56something ii a was something
7) Were all something
Oh and 1) F was twentysomething and G was 56 ?
i dont remember the answer .... but it was ... KE AT B = KE AT A + WDrGuys what was the answer to the very last question... it was something related to distance
WDr = AB * resistanceGuys what was the answer to the very last question... it was something related to distance
yeah !! all same but my length of string was about 1.66my answers
Q1> G=56
Q2> v=40m/s h=80m
Q6>co effiecient of friction = 0.571
Q?>length of string 1.42
t=1.066 a=0.6667
a=o at time t=40 and t=60 distance =440
ANYONE WITH SIMILAR ANSWERS?
what was ur at C ??? acceleration of ring ?? . gain in KE ?? and AB distance ??my answers
Q1> G=56
Q2> v=40m/s h=80m
Q6>co effiecient of friction = 0.571
Q?>length of string 1.42
t=1.066 a=0.6667
a=o at time t=40 and t=60 distance =440
ANYONE WITH SIMILAR ANSWERS?
I don't remember my change in k.e in q7 but my final speed was around 22 and initial was 13 or somethingwhat was ur at C ??? acceleration of ring ?? . gain in KE ?? and AB distance ??
The acceleration when it was attached to thei first found the speed of B when it reached which came out to be 1.871. (square root of 3.5) then used v^2=u^2+2as to find the distance P travelled and then subtracted that distance from 2.5 to get the length of string.
Um I got 1.94 , maybe because I rounded up some values so will I still get marks?I just solved it and I'm getting 1.9524m for my length of string.
T=1.0666...
a=2/3 (for 0.7m), then 2.8 (after the string breaks)
v (when string breaks) = *root* (2 x 2/3 x 0.7) = 0.96609
then it increases to 2
So
2^2 = 0.96609^2 + 2(2.8)s
s = 0.5476
So the length of string will be 2.5-0.5476 = 1.9523
yeaah. I got the same.in q2 v was 40 and h was 80 !!
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