Physics: Post your doubts here!

[autumnsakura]

Holy shit.I didnt understand this for days . Thank you so much ,

its okay. physic drives us mad sometimes

 

[autumnsakura]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_13.pdf
Q10

momentum is conserved so it should remain as mv
now, principleof conservation of momentum. mv= 2m(v) bracketed v is unknown.
By calculation, unknown v would be half the original v, v/2.
KE= 1/2 2m (v/2)2
observe here, it is 2m not just m.
So answer is A.

 

[queen of the legend]

Why does the height increase again ? Wont it be the same height. I agree the height increases by "h" in the right tube. but after that i dont understand.

originally the liquid's height was h ...when heated it expands and the increase is h again ...so its =2h

 

[EdmundH]

Then left tube will have height of x-h. and the right tube will have x + h right???? and now how come the difference is 2h ?

erm, difference = right - left = x + h - (x - h) = 2h
or either way which u get -2h (u just wan the magnitude of difference anyway)

 

[EdmundH]

Thank you so much! You are a saviour for dumb people like me !!

thx but dont ever say u're dumb! Good luck in ur exam!

 

[queen of the legend]

Can somebody post explainations to http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf

Questions 1, 27: the answers are C for both. I got the answer but messed up the power in both
Question 24C and 36B

for 27 : distance between one maximum and another is 15mm = 0.5 lambda
so lamda =0.03m
all electromagnetic waves travel at same speed= 3x 10^8 use v=f lambda , for f

f = 3 x 10^8 / 0.03
= 1 x 10^10 hz

for 24 : the amplitude of y is half of x by looking at the diagram= 4
now T= 1/F ...T for X = 1/100 = 0.01 s

Question 36 : if voltmeter reads 7.5 that means 7.5 volt is emf across the external resistor and 9-7.5 is lost in battery
since I=V/R , I=7.5/15 ..... I=0.5 amp
it takes one wave of x 0.01 s
if you count the number of waves in 0.01 s for y its 3 waves ...so t=0.01/3
Ty=3.3x10^-3 ......... and F=1/T .... F= 1/ 3.3x10^-3 , F=303 or approx. 300 hz , so answer is D

 

[queen of the legend]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf

^ Can someone please help me with questions 7,10,14,17,30,40,36
Thaankyou!

question 7 is direct application of formulas
for vertical displacement s= ut + o.5 gt^2 and horizontal is S=ut

question 10 : since f and m are directly proportinal ...heavier mass need more force to move .. so , masses given m and 3 m ...if F=1 N then 3/4 the force goes to 3m and 1/4th to m ( well this is how i could explain it )

q 14 : you need to focus on what question asks for
vertical component of force = 10000-9000=1000N

question 17 : take average current ...since its reduced over a period of time ...at one instant the current is not constant
so 100+20/2 = 60 x 8 = 480mC

question 36: earthed plate is considered positive so electric field is directed upwards and E=V/d (dnt forget to convert the units)

question 40 : here is a reply from one of the members (Taci12) :

Question 40:

According to syllabus, we must know that a proton has a charge +e (the basic/elementary charge)

Now according to this question, protons themselves are made up of particles called quarks. There are different types of quarks, each one with its own charge.

Whatever the combination of quarks, the total charge on a proton must be +e. Hence the sum of the charges of all the constituent quarks in a proton must be + e.

So simply check the sum of charges of each option to see if it is +e.

option A: 3 x (-e/3) = -e
option B: (2e/3) + (-e/3) = +(e/3)
option C: (2e/3) + 2(-e/3) = o
option D: 2(2e/3) + (-e/3) = +(3e/3) = +e

answer is D​

 

[Sriyog Sharma]

erm, difference = right - left = x + h - (x - h) = 2h
or either way which u get -2h (u just wan the magnitude of difference anyway)

thanks now i get it bro

 

[queen of the legend]

anyone knws how to solve this

22 A steel string on an electric guitar has the following properties.
diameter = 5.0 × 10–4
m
Young modulus = 2.0 × 1011
Pa
tension = 20 N
The string snaps, and contracts elastically.
By what percentage does a length l of a piece of the string contract?
A 5.1 × 10–4
% B 5.1 × 10–2
% C 1.3 × 10–4
% D 1.3 × 10–2
%

 

[emkay]

anyone knws how to solve this

22 A steel string on an electric guitar has the following properties.
diameter = 5.0 × 10–4
m
Young modulus = 2.0 × 1011
Pa
tension = 20 N
The string snaps, and contracts elastically.
By what percentage does a length l of a piece of the string contract?
A 5.1 × 10–4
% B 5.1 × 10–2
% C 1.3 × 10–4
% D 1.3 × 10–2
%

is the answer A. 5.1 * 10^-4?

 

[Silent Hunter]

anyone knws how to solve this

22 A steel string on an electric guitar has the following properties.
diameter = 5.0 × 10–4
m
Young modulus = 2.0 × 1011
Pa
tension = 20 N
The string snaps, and contracts elastically.
By what percentage does a length l of a piece of the string contract?
A 5.1 × 10–4
% B 5.1 × 10–2
% C 1.3 × 10–4
% D 1.3 × 10–2
%

is the answer B?

 

[geek101]

anyone knws how to solve this

22 A steel string on an electric guitar has the following properties.
diameter = 5.0 × 10–4
m
Young modulus = 2.0 × 1011
Pa
tension = 20 N
The string snaps, and contracts elastically.
By what percentage does a length l of a piece of the string contract?
A 5.1 × 10–4
% B 5.1 × 10–2
% C 1.3 × 10–4
% D 1.3 × 10–2
%

use the formula for E
E = F x L / delta L x A
rearrange to get the ratio delta L / L on one side (change in length / original length)
youll get >> delta L / L = F / E A (A = pi r ^2 )
= 20 / 2 x 10^11 x pi x (2.5 x 10^-4)^2
= 5.1 x 10^-4
multiply this by a 100 to get the %
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %

 

[emkay]

you party crasher!

use the formula for E
E = F x L / delta L x A
rearrange to get the ratio delta L / L on one side (change in length / original length)
youll get >> delta L / L = F / E A (A = pi r ^2 )
= 20 / 2 x 10^11 x pi x (2.5 x 10^-4)^2
= 5.1 x 10^-4
multiply this by a 100 to get the %
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %

why did you multiply it with 100 and made me look like a total idiot?

 

[geek101]

you party crasher!
why did you multiply it with 100 and made me look like a total idiot?

 

[Kumkum]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf
please help with question 12

 

[Kumkum]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf
question 17...please

 

[emkay]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
please help with question 12

the torque between the pulley Q and P in the upper wire is the torque at Q.
as it is causing the driving force for Q.
i.e T=F*d
F= T/d
F=3/100*10^-3
F=30*2 (since, the formula suggests force into perpendicular distance between the two forces)
F=60 N.
Torque on P,
T=F*d
T=30*150*10^-2 (Force remains the same, i.e 30N)
T=4.5 N m
(ps. Credit to geek101 for doing this)

 

[akshay.RONALDO]

i

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
question 17...please

i'm also stuck with this number. Please, do inbox me, if somebody replies.

 

[Kumkum]

the torque between the pulley Q and P in the upper wire is the torque at Q.
as it is causing the driving force for Q.
i.e T=F*d
F= T/d
F=3/100*10^-3
F=30*2 (since, the formula suggests force into perpendicular distance between the two forces)
F=60 N.
Torque on P,
T=F*d
T=30*150*10^-2 (Force remains the same, i.e 30N)
T=4.5 N m

thank you!

 

[Kumkum]

i
i'm also stuck with this number. Please, do inbox me, if somebody replies.

ya sure