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quite similar..though i scred up no.5..my answers are almost identical to yours. .
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quite similar..though i scred up no.5..my answers are almost identical to yours. .
gradient was 144+/-4 wasn't it?
And one of my % uncertainty came like 17.5% nd i wrote 18%, it it correct?
anyone plz write the answers of the second question
oh yes. right coz it's 1/v^2! Phew thank godThe gradient was positive, and in Q1, it WAS passing through the origin.
oh yes. right coz it's 1/v^2! Phew thank god
but er, it can't pass through the origin, coz when wind speed is 0, what are u gonna plot on the x-axis which reads 1/v^2 ? 1/0 is undefined..
also theta is 90 degrees, when v is 0, so tan 90 is again undefined. (0,0) cannot be a point, hence doesn't pass through the origin.
oh yes. right coz it's 1/v^2! Phew thank god
but er, it can't pass through the origin, coz when wind speed is 0, what are u gonna plot on the x-axis which reads 1/v^2 ? 1/0 is undefined..
also theta is 90 degrees, when v is 0, so tan 90 is again undefined. (0,0) cannot be a point, hence doesn't pass through the origin.
I think that a direct propotional graph is ONLY true if it is a STRAIGHT LINE PASSING THRU ORIGN! it is what we used throughout our practicals in As last year and also in O LEVELS too, lol!
And i looked up on the internet and it states tht if a graph has a y intercept it cannot b direct propotion :/!
Nd as tanQ was inverse prop. To v^2 hence for any value tanQ must be direct propotion to 1/v^2 hence a straight line passing thru orign shd b rite!! :/ otherwise tanQ cannot be inverse prop. To v^2 As was stated in the Q!
WAIT people ! we just have to write that if such garph was plotted and IF that was straight line passing thru origin thn the given relationship was CORRECT. Thats it.
Exactly thts what i wrote but actually it wont be like tht thts what im trying to xplain
TanQ was inversely proportional to v^2 so the equation we'd get would be tanQ=k/v^2. Comparing it with y=mx+c would give c and hence, the y-intercept to be 0. Doesn't that mean it should be passing through the origin? :/
no..it wouldn't pass through origin..it would if it was log smthin against smthin..
Wow!!! Log something against something!! U must b a genius to know that!!!
well if u think about it, the whole question of passing or not passing revolves around the concept of approximation.
1/v^2 will be o as v tends to infinity
and as v increases, the angle tends to infinity.
But well, u can't achieve a velocity of infinity with a table fan or whatever, hence you can't approximate 1/v^2 to zero, and thus tan theta to 0. It only tends to zero.
I still think it can't pass through origin
Thats something new there. You know what, **** it. Let the examiners decide. I'm outta here, gotta enjoy my holidats till they lastthe graph will be a curve NOT passing through the origin since it is not theta against v or theta against 1/v..its 1/v square...
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