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(a):
Thanks! But how do we get the no. Of maxima & minima?As u move along .. U know u see bright and dark fringes that is exactly what implies Here as well.. The dark fringes go for the minima and light go for the maxima .., in the same manner if u gi with the dots indicated .. U can say that If at O it is bright.. Then go along the dots and as u reach P is is maxima that is bright!
And u can say the readings go from maxima to minima ...and vice verse.
Hope it helps ya
Bro, could you explain the horizontal part again?it was 5 ms which is =1/4T
it was initially at the 1/4th from my diagram so add 1/4 we get 2/4th stage.
Oh, okay then. Thanks for all your help!
You guys confuse meBut the question didn't ask for drawing such diagrams ! I did this way and got the answer ! but I'm not sure did this answer 1 e (i) or not ! please someone correct me If I'm wrong !
Suchal Riaz
Thankyou so much!! <3a (i) So we know that the horizontal component is 8.2ms-1 since there the horizontal component never changes.
So Vcos60=8.2 (V is the Resultant velocity)
V=16.4ms-1
So, U know that vertical component =Vsin@
16.4sin60=14.2ms-1
ii) Since we are dealing with the vertical height here we will use the VERTICAL component of velocity.
So, initially u can see that vertical component of velocity=0
So v^2=u^2+2gh
(14.2)^2=2(9.81)h
h=10.3.
iii)This one is simple.U have to know the concept that the horizontal component of velocity remains constant.So the distance x u can easily get by multiplying the time taken with the horizontal component of velocity.So U find time like this:
V=u+gt
14.2=0+9.81t
t=1.4475s
X =v x t
X=8.2 x 1.4475
X=11.87m.
I hope u understood this question.If u have any problem i will try my best to respond.
Welcum.
Nope I think mine is right cuz I got the ans with it !You guys confuse me
what i did was just made a line continue the wind line but with a little less angle
i am not saying that you are wrong but i don't understand the way you are going it
Since, 5 ms is 1/4th of total T, we draw a straight line?
Maxima is the bright fringes and minima is the dark. Thus, u can count.
Hope this help uThey are saying that point A moves 80mm. Point A is a point on the rope and it is not moving sideways but only moving up and down.
So it moves 80mm first going up to amplitude then it goes back to equilibrium position back down to a trough and back up so the actual amplitude is 80/4=20.
DeViL gURl B)
GImme the paper link please.
It came once in M/J 2002 not in Xtreme Papers .. you'll have to go for www.freeexampapers.net
Link not avilable :/
since the wave is already at 1/4 now by adding 1/4 it becomes 2/4 so it's a horizontal line
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