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AS Physics P1 MCQs Preparation Thread.

omg

omg

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I NEED HELP IN QS. :
14
15
27
AND
34!!!!!!!!!!!!!!!!!!!
PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS :(((
 

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N.M

N.M

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SalmanslK said:
View attachment 12388

the answers B. But how? any description will be appreciated.thank you.

and can anyone tell the effects of changing any one one value will have on other values in d.sin(teta)=n.lamda
Click to expand...

by increasing the number of lines, the distance between the lines decreases
u can also understand by this formula
d=1/N
where d is the distance b/w the lines
N is the number of lines

So now we know that the distance b/w the lines is dec
according to the formula
dsin@=n(lamda)
if u dec d that means that the angle will inc b/c it is inersely proportional to d
similarily, as d dec, the value of n also dec becuz they are directly proportional

hope it is clear now
 
A.ELWY 7

A.ELWY 7

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omg said:
I NEED HELP IN QS. :
14
15
27
AND
34!!!!!!!!!!!!!!!!!!!
PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS :(((
Click to expand...
I told u I promise..ok, now...
Q15: first u need to get the angle theta of the triangle using...opp/hyp=sin ( theta) so 3/7 = sin ( theta) so sin^-1 ( 3/7) = 25.4...then when resolving forces..the force down the inclined plane is mg sin ( theta) so 2 x 9.81 sin ( 25.4) = 8.42...then 8.42 - 5n ( 5 N is the friction) so 3.42N then...F=ma so 3.4 /2 = a = 1.7...finally V^2= 2as...so V= square root of 2 x 1.7 x 7= 4.9 so A
27: 500 lines per metre..so to get d..d = (1/500) x 10^-3...( i multiplied by 10^-3 cause it is mm) so ull get 2 x 10^-6 then d x sin( theta) = n ( lambda) so 2 x 10^-6 sin ( 90) = n x ( 600 x 10^-9)..note( sin 90 means to observe the whole grating ..and 600 x 10^-9 because it wqas nano metre)..so n=3.3 so 3....then he asked for the images..here we got n which is the number of the maximum level..so we have the normal light pasin straight so this is 1..and we have 3 above the normal and 3 down the normal so the total is 7.....note( in the future..if he asked for the images or the beams the observer will see, and u got n from the calculation..use the equation 2n +1) if we used it here we will get 7...as (2 x 3) + 1 = 7
34: resistance is equal to = pl/A and the volume is Ah...we have and h which is the l...the area of a cube is L1 x L2 so pl/L1 x L2 and the volume of a cube is L1 x L2 x L3 so it is, when substituting.....p/V^1/3 so C
 
omg

omg

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A.ELWY 7 said:
I told u I promise..ok, now...
Q15: first u need to get the angle theta of the triangle using...opp/hyp=sin ( theta) so 3/7 = sin ( theta) so sin^-1 ( 3/7) = 25.4...then when resolving forces..the force down the inclined plane is mg sin ( theta) so 2 x 9.81 sin ( 25.4) = 8.42...then 8.42 - 5n ( 5 N is the friction) so 3.42N then...F=ma so 3.4 /2 = a = 1.7...finally V^2= 2as...so V= square root of 2 x 1.7 x 7= 4.9 so A
27: 500 lines per metre..so to get d..d = (1/500) x 10^-3...( i multiplied by 10^-3 cause it is mm) so ull get 2 x 10^-6 then d x sin( theta) = n ( lambda) so 2 x 10^-6 sin ( 90) = n x ( 600 x 10^-9)..note( sin 90 means to observe the whole grating ..and 600 x 10^-9 because it wqas nano metre)..so n=3.3 so 3....then he asked for the images..here we got n which is the number of the maximum level..so we have the normal light pasin straight so this is 1..and we have 3 above the normal and 3 down the normal so the total is 7.....note( in the future..if he asked for the images or the beams the observer will see, and u got n from the calculation..use the equation 2n +1) if we used it here we will get 7...as (2 x 3) + 1 = 7
34: resistance is equal to = pl/A and the volume is Ah...we have and h which is the l...the area of a cube is L1 x L2 so pl/L1 x L2 and the volume of a cube is L1 x L2 x L3 so it is, when substituting.....p/V^1/3 so C
Click to expand...

for qs 34 can u help me do the substitution :/
 
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bloody_mary said:
please help me with questions:
15
26
28
30
31

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_12.pdf
Click to expand...


15 - W= FxD
8-4 =4
4= what values of F and s
so therefore assume F=2 AND s=2
no double both F and S as said in the Q.
W= 4x4 = 16
16 + initial 4 = 20J

-28 ) White light source emits light at same frequence. plus all other options are absolutely wrong.
-30) field lines go from + to -
therefore electron deflected upwards at greater angle because they have small mass.
 
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