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Question 33 same paperWhich question ._. ??
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Question 33 same paperWhich question ._. ??
Dude, in question 5 .. I got till 0.02 seconds .. But then y did u divide by 2?5)
![]()
---------->Written form :
Frequency: 3000 oscillations in a minute so , 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01
7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory
9)Use v^2 - u^2 = 2as
v has to be zero, since the train has to come to rest eventually.
s = x
The equation becomes: 2ax + u^2 = 0
make 's' distance the subject:
x = - (u^2)/2a
deceleration, i.e -a, is constant in the question.
take the constants out,
you get this relation: x ∝ u^2, i.e, s varies as the square of u.
So when u is increased by 20% (120/100 = 1.2), x will change by (1.20)^2 = 1.44x
Ahh... I'm afraid nahi samajh aai..It has to be done like this :¬ Hope shamajh ajaye
View attachment 43646
Can we use the logic here that when they say area then it's R and when they day radius it's R/2 ?It has to be done like this :¬ Hope shamajh ajaye
View attachment 43646
Read last part in pic...Dude, in question 5 .. I got till 0.02 seconds .. But then y did u divide by 2?
And then when you got 0.01 how's that supposed to make up with 10 in the answer given :$
Clear nai hainAhh... I'm afraid nahi samajh aai..
Thanks for the help though
A thora sa clear view would do the job![]()
Hindi likho, English weak hain meriCan we use the logic here that when they say area then it's R and when they day radius it's R/2 ?![]()
Dude where do u get 120 from in the last question u explained ? :$5)
![]()
---------->Written form :
Frequency: 3000 oscillations in a minute so , 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01
7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory
9)Use v^2 - u^2 = 2as
v has to be zero, since the train has to come to rest eventually.
s = x
The equation becomes: 2ax + u^2 = 0
make 's' distance the subject:
x = - (u^2)/2a
deceleration, i.e -a, is constant in the question.
take the constants out,
you get this relation: x ∝ u^2, i.e, s varies as the square of u.
So when u is increased by 20% (120/100 = 1.2), x will change by (1.20)^2 = 1.44x
Okhay.. Like Kia hum only R likh saktae hain when they say area and only R/2 likh saktae hain when they say Radius ? Samajh aai?Hindi likho, English weak hain meri![]()
Nope..Clear nai hain?
Well though I'm an AS student but .. Know one thing that u do get marks for the right concept as in I dunno what to I call it .. There's a name for that but.. U do get marks for usin the right concept sisI'm really sorry if anyone of you get disturbed reading my message here. Good to see lot many friends working hard for paper 1 (ma sha Allah).
I have a wired doubt, its about physics p5. Yes, don't we lose the entire 15 marks in the first question if we write the independent variable wrong cause I did that mistake and wrote the whole concept wrong
(already both p2 & p4 went bad although p2 was easy)
5)
9) First find the time, using S = Ut + 0.5t^(2) ; S = 1.25 , U = zero, t = ?
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