PHYSICS P2 HELP

[ninjas4life]

can anyone explain to me Q6 (b) [electricity question] ??
from the oct/nov 2010 physics paper 22
http://www.xtremepapers.me/CIE/inde...Level/9702 - Physics/&file=9702_w10_qp_22.pdf

if anyone has any physics doubts you can also get it posted and maybe get help here!!

 

[MHHaider707]

can anyone explain to me Q6 (b) [electricity question] ??
from the oct/nov 2010 physics paper 22
http://www.xtremepapers.me/CIE/inde...Level/9702 - Physics/&file=9702_w10_qp_22.pdf

if anyone has any physics doubts you can also get it posted and maybe get help here!!

1) across A and B voltage is 4V so it means that across 1.2k(ohm) resister, voltage is (9-4)= 5V

calculate current by V = IR
I = 5/1200

Total current in 1.6k(ohm) resister and thermistor is (5/1200)

So again by V= IR
R = 4/(5?1200) =960 (ohm)


2) total effective resistance across A and B is 960 (ohm)

1.6k(ohm) resister and thermistor are in parallel so calculate resistance of thermistor by

(1/960)-(1/1600) = (1/x) :: x is resistance of thermistor,

x = 2.4 k(ohm)

read the value of temperature from graph and it is 11*C


Hope it helped

 

[Hateexams93]

PEOPLE can some1 plz do may june 2009 Q5 (b)???????????

 

[MHHaider707]

PEOPLE can some1 plz do may june 2009 Q5 (b)???????????

First we have to calculate path difference....
Distance between S2 and M is 128cm (calculate it by pythagoras theorem)

path difference is 128-100 = 28cm

when frequency is increased from 1000 Hz to 4000 Hz...wavelength decreases so we calculate the range for wavelength by v=f(lemda)
At f= 1000Hz, wavelength = (330/1000) = 0.33m
At f= 4000Hz, wavelength = (330/4000) = 0.0825m

we know that, for minima path difference = 1/2(lemda) , 3/2(lemda) , 5/2(lemda)

so equate 1/2(lemda) = path difference
wavelength = 0.56m (it is not in the range, so reject it)

3/2(lemda) = path difference
wavelength = 0.19m (it is in the range so first minimum)

5/2(lemda) = pathdifference
wavelength = 11.2m (it is again in the range so this is econd minimum)

next minimum is not in the range 0.33<lemda>0.0825

so only two minima are detected

 

[Hateexams93]

so the range is between min and max lemda ?

 

[MHHaider707]

so the range is between min and max lemda ?

yeah, and it is calculated by the given frequency values i_e 1000Hz - 4000Hz


Ifb you don' t understand it . . . post your Question in this thread:

viewtopic.php?f=26&t=7477

 

[Hateexams93]

hey can u plz help me with 1 more question , october november 2005 4 (a) ii

 

[MHHaider707]

hey can u plz help me with 1 more question , october november 2005 4 (a) ii

you can find the area under the graph by drawing tangents to the curve and try to take maximum of the area......

 

[Hateexams93]

i tried , it didn't work with me

 

[MHHaider707]

After every 0.05s, take area under the graph separately and add them all.....

 

[Hateexams93]

can i use integration ?

 

[nidzzz09]

Wats the difference between progressive and stationary wave ?

 

[princesszahra]

here u go!

 

[nidzzz09]

here u go!

Thanks alot but i dont understand the the first three points Can you please explain

Lke wats do you mean when yu say the amplitude of the vibration varies with position along the string: its zero at the nodes and maximum at an antinode . in progressive waves all the points have the same amplitude :x

 

[MHHaider707]

can i use integration ?

Yeah may be :%)

 

[shyqueen]

plz sumbodyyy explain answer to Q6 (b)(iii) nov10/p21....!!!!