PHYICS!

[libra94]

Assalaomo Alaikum.

http://www.xtremepapers.net/CIE/Interna ... 8_qp_2.pdf
Q 3(c)i)

http://www.xtremepapers.net/CIE/Interna ... 4_qp_1.pdf
Q 25

thankyou.

 

[libra94]

Re: PHYSICS!

anyone???

 

[usman]

http://www.xtremepapers.net/CIE/Interna ... 8_qp_2.pdf
Q 3(c)(i):-
This question is typical of M1 (Maths) "Force Resolution" topic. U cn solve this particular question by realizing that the weight of the trolley can be resolved in two perpendicular directions: one parallel to the slope while the other normal to it. In this way, u'll find that the component down the slope of the trolley’s weight is: W(normal to slope) = 42*g(=9.81)*cos(2.8deg) while W(parallel to slope) = 42*g(=9.81)*sin(2.8deg). It is the W(parallel to slope) which has been asked in the question

 

[usman]

http://www.xtremepapers.net/CIE/Interna ... 4_qp_1.pdf
Q 25
As far as this question is concerned, i can tell u what method i personally employ to tackle this type of question. It is this:-
1. U need to know the direction in which the wave is heading
2. An arbitrary particle on the wave will take the position of the particle infinitesimally/ri8 before it. In other words, the waveform will shift in the direction in which the wave is travelling
Applying this info to the given question:-
It is almost obvious that the particle P is moving downwards (since it is going to take the position of the particle behind it, which is at a reasonably lower than P). But the case for the particle Q is a lill different, it is at a peak(crest in this case) i.e., the particle infinitesimally left of it is almost the same level as itself. This implies that, at this very instant, it ain't moving at all

M afraid, i hav confused u with such an inefficient method of mine. Bt anyways, I hop this helps