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thxx


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thnk ulet p's time = t
then q's time = t - 2
a = 1.75 and u = 0 for both
Displacement of P = 0.875t^2
Displacement of Q = 0.875(t-2)^2
Displacement of P - Displacement of Q = 4.9
0.875t^2 - 0.875(t-2)^2 = 4.9
t = 2.4
but since it's talking about time in terms of Q in the question
t = 2.4 - 2
= 0.4
The 2 parts of the string are at different angles.http://maxpapers.com/wp-content/uploads/2012/11/9709_s12_qp_41.pdf
For question 7, why can't I assume that the tensions on both parts of string (BC and BA) are same? I mean the string is not attached to any point between A and C, it is only pulled by a force of 8N. Should i assume that tension is related to length of string between 8N force ??
Pleease Help??
thx
They said how far it can go so that means final velocity will be 0. And it has negative acceleration so that also tells u that it is slowing downthx
but in the ii part how did u now that v =0
can some one help please.. im really confused on when to take the weight component of an object into consideration because sometimes i use it where im nt supposed to and sometimes i dont use where im supposed to use it. For example I didnt know that we need to use it in s14 p43 Q2 (what i mean by the weight component is the gravity force exerted on the mass of the object at an inclined plane which is usually calculated by mgSin(Theta) )
Always use it when you are using newton's second law ( Fnet = ma)can some one help please.. im really confused on when to take the weight component of an object into consideration because sometimes i use it where im nt supposed to and sometimes i dont use where im supposed to use it. For example I didnt know that we need to use it in s14 p43 Q2 (what i mean by the weight component is the gravity force exerted on the mass of the object at an inclined plane which is usually calculated by mgSin(Theta) )
i)V = P/FView attachment 53489 Q3 explanation please?
thank you bhaiAlways use it when you are using newton's second law ( Fnet = ma)
i)V = P/F
If P increases, V increases, so Vb = 1.2* 28
If F decreases then V increases, so Vb = (1.2 * 28)/0.96 (multiplying by 0.96 will decrease V, so it's divided)
ii)Work done by Driving force + Loss = Work done by resistive force + gain
WD + 0 = (2.3 * 10^6) + (.5)(200000)(35^2-28^2)
can some one help please.. im really confused on when to take the weight component of an object into consideration because sometimes i use it where im nt supposed to and sometimes i dont use where im supposed to use it. For example I didnt know that we need to use it in s14 p43 Q2 (what i mean by the weight component is the gravity force exerted on the mass of the object at an inclined plane which is usually calculated by mgSin(Theta) )
graph plz
Tension on both sides of the pulley is same, so the right angle can be split into 2 45 degree angles. If you resolve one of these, it's Tcos45.P4 MJ 06 Q5 (i) .... I SERIOUSLY DON'T GET IT! -freaking out-
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