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I dont think so, as it would be part of the Unit 6 (part a i mean) but part b can be either
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Physics Unit 4 JUNE 11/6/12
- Thread starter Evolution
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yaI dont think so, as it would be part of the Unit 6 (part a i mean) but part b can be either
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
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Sure no problem n dont apologize pleaseya
Its practice for me too Just give me a while, m still revising the chapter
ya sureSure no problem n dont apologize please
Just give me a while, m still revising the chapter
and btw u wil revise it frm the book or revision guide?
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The revision guideya sure
btw for Q4 ii, I checked the mark scheme, and the formula they r using is new to me
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 checkThe revision guide
btw for Q4 ii, I checked the mark scheme, and the formula they r using is new to me
But u can do it using Gradient of the steep straight part
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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?
Never seen it before! 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)ohh.. but how is the v maz at t=0? :S Souldnt v be 0 at that point? :S
and which formula is this?
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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 isnt v max when h =0? :S
ohhh den i used this formula too
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 correcti 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
ohhh sry t wil nt be zero according 2 the graph i wil do the que again
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i'm confused y ur saying dat t=0 when when displacement is 0, if u see this particular graph, i think its at 0.32sya 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
dw, m not sure on this myself
cuz my answer comes 1.7
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No problemohhh sry t wil nt be zero according 2 the graph i wil do the que again
But i'm not getting the answer :S I'll try again... still i didn study astrophysics i didn finish anythng til nwNo problem
im affraid if time won't fit me
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SAME HEREEEEEEEEEEEEEEEEstill i didn study astrophysics i didn finish anythng til nw
I have to do bio as well and i just started physics -____-"BUT we'll manage SOMEhow Inshallah dw
ya sure inshallaSAME HEREEEEEEEEEEEEEEEE
BUT we'll manage SOMEhow Inshallah dw
just 1 more last queSAME HEREEEEEEEEEEEEEEEE
BUT we'll manage SOMEhow Inshallah dw
in june 10 last part of the last que
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Keep em comming no problemjust 1 more last que
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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Can u please the paper cuz i dont have itjust 1 more last que
last part of last que and 4 que (4) i got stuck in amplitudeKeep em comming no problem
did u get the answer for 4? I did the rest of that question, what did u want me to explain again?