Physics P2 Help

[flyingcrayons]

November 2007 Q5(a) - What is the right way to sketch the graph? The marking scheme says 3 peaks. My graph only has 2 peaks. What am I doing wrong? Help is appreciated! Thank you in advance!

 

[flyingcrayons]

For part (c)(i), why are fewer fringes observed?

 

[flyingcrayons]

Anyone?

 

[MAVtKnmJ]

It says "3 Peaks" which includes positive and negative peaks. If you have drawn this correctly, you will get two positive peaks and one negative peak at the center.

There are fewer fringes now because the rays will fall on a smaller area now. It's like when you have a flash light and you aim it on a wall you will see that it will light up a larger area of the wall. But if you cover the top and bottom halves of the torch, then you will notice that now the torch is lighting up a smaller area on the wall.

This is not the best or most accurate example but it should get the point through to you.

 

[beacon_of_light]

It says "3 Peaks" which includes positive and negative peaks. If you have drawn this correctly, you will get two positive peaks and one negative peak at the center.

There are fewer fringes now because the rays will fall on a smaller area now. It's like when you have a flash light and you aim it on a wall you will see that it will light up a larger area of the wall. But if you cover the top and bottom halves of the torch, then you will notice that now the torch is lighting up a smaller area on the wall.

This is not the best or most accurate example but it should get the point through to you.

yea MAVtKnmj is right....but a lil more explanation

find the phase difference in terms of seconds so that will be easier to draw a wave having 3 peaks...lets consider 30s for the wave drawn as its time period so the new wave wuld have the same time period too...as using the formula calculate phase diff....

x/30 = 60/360

x=5s...which means every point on the new wave wuld be 5 s ..behind the already drawn wave...start the wave at t = 5s which is 5 divisions and carry one...hope u gt that........

 

[beacon_of_light]

and for few fringes...when width of each slit increases....diffraction reduces...and when there is less diffraction less fringes r observed...eg we get larger no of fringes using diffraction grating thn when using 2 slits only....

 

[flyingcrayons]

Got it! Thanks!

 

[flyingcrayons]

June 2005, can someone please explain Q7 part (c)? I don't get it. Thanks!

 

[beacon_of_light]

june05:
P =IV.............here in the circuit 'I' would be same in R and r since both are connected in series....

So it means power dissipation is directly proportional to V....
At 1.12V around R, voltage around r would be 3-1.12=1.88V...which means higher pd across r results in larger power dissipation....

At 1.9V around R, voltage around r would be 3-1.9=1.1V....so pd around r is less which means less power dissipation in r........So at a higher pd value across''R''.....pd across ''r'' would be low and hence power dissipation too....

 

[flyingcrayons]

Thanks!!! What about November 2004, Q6 (c)? Why is p.d. across C larger than across R? (according to marking scheme)

 

[aleem420]

Can anyone help me understand M/J/2009/21 Q5(b) quickly as possible ..................

 

[Jurol]

well, this question was a pain!

since s1 and s2 are coherent, and they're asking us to find the minima, which means odd multiples of π
so, you take phase difference = (2n+1)π for the minima okay, and then we know that phase difference is also equal to phase difference= (2π/λ)x , where x is the path difference and the path difference we can calculate from the diagram, by taking the hypotenuse from S2 and M, we get 128, then 128-100, coz the path difference from M then we get 28.
equate these 2 formulas (2π/λ)x=(2n+1)π - we get λ by using v=fλ then, λ-v/f - where f is 4kHz-1kHz which is 3kHz okay, then

then, π gets cancelled from both the sides,

(2/(330/3000))x28/100 = (2n+1) --- 28/100 coz its in cms we need it in meters
(2/0.11)x28/100=(2n+1)
n=2.045454545
therefore n = two minma!

Over and Out
Jurol

 

[flyingcrayons]

Bump

Thanks!!! What about November 2004, Q6 (c)? Why is p.d. across C larger than across R? (according to marking scheme)

 

[flyingcrayons]

Nov 2003 Q5 (c) - is the graph supposed to curve upwards?

 

[flyingcrayons]

Anyone???? Please?

 

[Azazel]

Yes, it'll curve because the electron starts from rest.

as for W04, R = V/I

if V increases, R increases too.