CIE physics paper 2

[Deadly_king]

Thank you..I am flattered I like that I encourage to help

Hehe........i hope that 1 day i may be as encouraging as you are to others

[Greed444]

can someone help explain what is phase angle 60 degree as in the OCT/NOV 2002 Q5
it would be great if someone can show me how to do the Q5b(i) too

[Deadly_king]

can someone help explain what is phase angle 60 degree as in the OCT/NOV 2002 Q5
it would be great if someone can show me how to do the Q5b(i) too

By a phase angle of 60 degrees.......it meant that the wave T₂ would be moving 60 degrees faster.

In other words, when wave T₁ is at maximum amplitude at time t=0, wave T₂ will be at maximum amplitude at time t=0.5

But both waves will be having same frequency, wavelength and period.
60 degrees is the phase angle. To be able to do Q5b(i) you should convert the phase angle to the time difference sine the graph is one of X against time.
Phase angle = (delta T)/T * 360

From the graph it is noted that period T= 3s.
Hence delta T will be 0.5s.

You just need to draw a wave with same amplitude, same frequency and wavelength........but which is just 0.5s faster.

Hope you understood........if not let me know and i'll try to be clearer

[S.M.A.T]

Keep it up DK


+rep

[Greed444]

By a phase angle of 60 degrees.......it meant that the wave T₂ would be moving 60 degrees faster.

In other words, when wave T₁ is at maximum amplitude at time t=0, wave T₂ will be at maximum amplitude at time t=0.5

But both waves will be having same frequency, wavelength and period.
60 degrees is the phase angle. To be able to do Q5b(i) you should convert the phase angle to the time difference sine the graph is one of X against time.
Phase angle = (delta T)/T * 360

From the graph it is noted that period T= 3s.
Hence delta T will be 0.5s.

You just need to draw a wave with same amplitude, same frequency and wavelength........but which is just 0.5s faster.

Hope you understood........if not let me know and i'll try to be clearer

Thank you DK ^^ your explanations are crystal clear!

[Deadly_king]

You are welcome

Glad to have been able to clear your doubts 

Thanks Asif 

[Greed444]

i have a problem on O/N 2004 paper 2
its on Q6b (ii) & c

[S.M.A.T]

b ii)Potential difference acrooss C&R are same as emf(because of parallel connection)

From the graph:
when v=2V
current in R=1.3A
current in C=0.75A
therefore,current in the battery+1.3+0.75=2.05A
c)The current is same in both R&C as they are connected in series.But the potential difference across C is higher than R.Since power=VI,therefore component c will dissipate thermal energy at a greater rate

[Greed444]

many thanks asif!

[S.M.A.T]

U WELCOME