dnt mind giving me the link where i can check the threshold for all papersfor paper 22 that's right......
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dnt mind giving me the link where i can check the threshold for all papersfor paper 22 that's right......
force cant be on the y axisaxes was definitely force on y axis and extension on x axis, still confused about the end point.
distance between 2 nodes is half lambdadistance between two nodes IS lambda not lamba into 2
force cant be on the y axis
in that experiment force is the independent variable which means we change force and extension is dependent variable which means extension varies with force
So it should be Force on x axis and Extension on y axis.
If this is so the second graph should have twice the gradient of the graph already drawn
I am 100% sure that force was on y-axis!!!! And the gradient would have been halved.force cant be on the y axis
in that experiment force is the independent variable which means we change force and extension is dependent variable which means extension varies with force
So it should be Force on x axis and Extension on y axis.
If this is so the second graph should have twice the gradient of the graph already drawn
Force was on y-axis!!I am 100% sure that force was on y-axis!!!! And the gradient would have been halved.
xactlyForce was on y-axis!!
buddy its wrong im 100% SUREdude its strain on the y axis and stress on the x-axis
read the previous post. i apologized. my concepts r clear as its obvious by my answer of 500hz. it was a simple typing mistakeLOL. I would suggest you to revise your concepts before arguing.
if force was on y axis then gradient of the graph for copper is half the gradient of that for steel.hey, i get your point..but the question tht day was force on the y-axis n extension on x-axis..the original graph, the end point was at 3N (y.axis) and 0.25mm (x-axis)..remember?? so, how the graph should be??
read the previous post. i apologized. my concepts r clear as its obvious by my answer of 500hz. it was a simple typing mistake
the guys who got 1000 are the idiots500 Hz was NOT the answer, I repeat. The idiots who got 500 probably made the mistake of thinking it was a uniform stationary wave, taking the distance from the start to the end of the wave to be 0.68m and the distance between three nodes. Lol, I say. It was a PART of the wave, and so the distance between two successive nodes/antinodes somewhere in the middle of the wave had to be taken.
the guys who got 1000 are the idiots
YepLet's just wait for the mark scheme then.
in a part of the wave distance between 3 nodes is the wavelength. even if u took distance between two nodes u had to multiply it by 2500 Hz was NOT the answer, I repeat. The idiots who got 500 probably made the mistake of thinking it was a uniform stationary wave, taking the distance from the start to the end of the wave to be 0.68m and the distance between three nodes. Lol, I say. It was a PART of the wave, and so the distance between two successive nodes/antinodes somewhere in the middle of the wave had to be taken.
in a part of the wave distance between 3 nodes is the wavelength. even if u took distance between two nodes u had to multiply it by 2
Okay i got this question wrong but i solved it afterwards and the ans was 500Hz how? let me explain read this properly and please everyone SHUT the F___ UP..Yes, exactly. And if I remember correctly, the distance between two successive antinodes was 17cm=0.17m. So, 0.17x2=0.34m. And hence, 340/0.34=1000Hz.
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