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AS Physics P1 MCQs Preparation Thread.

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

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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

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

Messages
626
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3,628
Points
253
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

for qs 34 can u help me do the substitution :/
 
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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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