AS Physics P1 MCQs Preparation Thread.

[omg]

remove 3 and sin from the eq has they have no units... this gives you L(a^2/P)=QT^2
now put their units.. m(m^2/P)=Qs^2.. by solving this, u will have QP=m^3/s^2... or m^3s^-2.... now search through the answers whose product can give you m^3s^-2.. and its B.. (m^2) * (ms^-2)

ooooooo.. thankyewwwwwwwwwwwwwwwwwwwww

 

[bloody_mary]

Could anyone answer the question in the attached image? The answer is A but I don't get it.

electric fields always travel from + to -
so speed will decrease as it electron is attracted to + not negative
path is linear because it is travellin parallel to electric field not perpendicular to cause parabolic motion

 

[Mattman]

Could anyone answer the question in the attached image? The answer is A but I don't get it.

The path of E is parallel to velocity so theres no reason for it to be parabolic. since E is pointing to the right, the positive area is behind the electron. So, the electron is being pulled back, hence decelerating.

 

[umarashraf]

ooooooo.. thankyewwwwwwwwwwwwwwwwwwwww

welcome jee.. duaon mein yaad rakhna bs..

 

[USMAN Sheikh]

lesco thanks alot man u r doing a great job few more questions for u kindly solve them pls
q 14 and 36 of june 11 variant 11
q 26 28 36 of november 11 variant 11

hELP ME OUT WITH DESE PLS posted on page 38 pls

 

[raamish]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf

qs 9) tell how to do it

 

[MysteRyGiRl]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
Q no 29...why do we use cos?? :S
Q no 36 explanation needed plz..

 

[mariyam95]

thanx

 

[bloody_mary]

hELP ME OUT WITH DESE PLS posted on page 38 pls

for Q14 i have no clue i am stuck on it myself

Q36 well as you can see Wand X are in parallel so the total resistance of both of them is less than resistance of one them if it is in series

since the whole circuit is in series current is constant so P=I^2 R if R goes up P goes up
So now there is a higher resistance in X so more power

------------------
for Q26 ▲l is the bottom length as this is the one hat decreases with lenght but as it decreases the other L increases because the cube is flattened so they are inversely proportional

Q28 don't know why :S
Q36
V decreases because already the voltage across the battery is V including internal resistor adding another resistor will require more energy so V will decrease

 

[bloody_mary]

can someone 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

 

[i love my country]

plz i need help in M\J 2007 Q8,12,20 ....

 

[i love my country]

OH SORRY AND Q 3

 

[omg]

welcome jee.. duaon mein yaad rakhna bs..

zarorrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

 

[MysteRyGiRl]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
Q no 29...why do we use cos?? :S
Q no 36 explanation needed plz..

ppl help plzzz

 

[USMAN Sheikh]

welcome jee.. duaon mein yaad rakhna bs..

hELP ME OUT WITH DESE PLS posted on page 38 pls

plsumar can u help me out with dese with details

 

[omg]

ppl help plzzz

for q.36 as the resistors are in series u cn apply the potential divider formula : (180/180+120+150) x 6

 

[MysteRyGiRl]

for q.36 as the resistors are in series u cn apply the potential divider formula : (180/180+120+150) x 6

ohh thnxx...can u explain Q no 29 too?

 

[xyz!]

plz i need help in M\J 2007 Q8,12,20 ....

Q.8:-
c here u cant directly find out the time taken for last 10m so u got to find out the time taken for 40m and for the first 30m separately and then subtract the 2...
consider s=40m first, so u=o, a=9.81
so using s=ut+0.5at^2
we get t=2.855s
Then considr s=30m so again using the same formula u get t=2.47s
subtracting the two gives u t=0.38s


Q.12:-
p(b4 collision)= (4*2)+(1*4)=12
p(after collision)=6*v=6v
p(b4 collision)=p(after collision)
so 12=6v; v=2m/s
now find out the k.e after collision: 0.5mv^2= 0.5(6)(2^2)=12J==> B

 

[USMAN Sheikh]

HELP ME ON THESE QUESTIONS PLS
15 22 32 of n10 variant 12
16 19 34 of J11 variant 12

waiting for quick reply pls with details on this and my previous asked questions on page 38 pls lesco where are u

 

[omg]

ohh thnxx...can u explain Q no 29 too?

idkk abt that qs, m left wid all these diffraction grating qs. will do dm IA tonyt and maybe tomorrow i cn help u