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Physics Unit 4 JUNE 11/6/12

How well did you do on the exam?


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ya :) ok i ned help also 4 q(12)last part and q(14) whn they ask abt verification tht rope won't slack :)and sry again
Sure no problem n dont apologize please ;) Its practice for me too :D

Just give me a while, m still revising the chapter :oops:
 
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ya sure :)and btw u wil revise it frm the book or revision guide?
The revision guide :)

btw for Q4 ii, I checked the mark scheme, and the formula they r using is new to me :eek:
But u can do it using Gradient of the steep straight part
 
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The revision guide :)

btw for Q4 ii, I checked the mark scheme, and the formula they r using is new to me :eek:
But u can do it using Gradient of the steep straight part
actually i did it by taking time as 0 when max velocity is v=Ax(angular velocity)do ths and check
 
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actually i did it by taking time as 0 when max velocity is v=Ax(angular velocity)do ths and check
ohh.. but how is the v maz at t=0? :S Souldnt v be 0 at that point? :S
and which formula is this? :eek: Never seen it before! :eek:
 
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ohh.. but how is the v maz at t=0? :S Souldnt v be 0 at that point? :S
and which formula is this? :eek: Never seen it before! :eek:
v is max whn it is in the equilibrium position so i guess its zero time and displacement is also zero and its the same formula for velocity bt cuz (A x angular velocity x sin x angular velocity x time is zero so A x angular velocity is left)
 
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v is max whn it is in the equilibrium position so i guess its zero time and displacement is also zero and its the same formula for velocity bt cuz (A x angular velocity x sin x angular velocity x time is zero so A x angular velocity is left)
i still dont get it, at t=0 the velocity is max according to the graph right? :S so the gradient at that point (that is the velocity) will be 0 isnt it? :S isnt v max when h =0? :S
ohhh den i used this formula too :p
 
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i still dont get it, at t=0 the velocity is max according to the graph right? :S so the gradient at that point (that is the velocity) will be 0 isnt it? :S
ohhh den i used this formula too :p
ya i kno tht whn they say max velocity it means displacement is zero hence also time wil be zero at equilibrium position ths wat i kno i don kno if its correct:(
 
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v is max whn it is in the equilibrium position so i guess its zero time and displacement is also zero and its the same formula for velocity bt cuz (A x angular velocity x sin x angular velocity x time is zero so A x angular velocity is left)
ohhh sry t wil nt be zero according 2 the graph i wil do the que again :)
 
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ya i kno tht whn they say max velocity it means displacement is zero hence also time wil be zero at equilibrium position ths wat i kno i don kno if its correct:(
i'm confused y ur saying dat t=0 when when displacement is 0, if u see this particular graph, i think its at 0.32s :confused:
dw, m not sure on this myself :unsure: cuz my answer comes 1.7 :eek:
 
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still i didn study astrophysics i didn finish anythng til nw:(im affraid if time won't fit me
SAME HEREEEEEEEEEEEEEEEE :cry: :cry: :cry: I have to do bio as well :( and i just started physics -____-"
BUT we'll manage SOMEhow Inshallah dw :)
 
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just 1 more last que :) in june 10 last part of the last que
Keep em comming no problem ;) n just 1 sec, i'll check dat...

did u get the answer for 4? I did the rest of that question, what did u want me to explain again? :)
 
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Keep em comming no problem ;) n just 1 sec, i'll check dat...

did u get the answer for 4? I did the rest of that question, what did u want me to explain again? :)
last part of last que and 4 que (4) i got stuck in amplitude
 

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