Physics: Post your doubts here!

[Dr. Seuss]

how do you do these questions ?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

questions 7, 14, 15, 16, 27, 30, 31, 36, and 37

 

[Thought blocker]

5,9,15,24,30,31,33,34 help pleasee

5)
We know S = d / t = 40 / 2.5 = 16 m/s
Uncertainty in speed = (uncertainty in distance + uncertainty in time) x speed
--------"--------------- = ( (0.1/40 ) + (0.05/2.5) ) x 16 = 0.36 ≈ 0.4 (1dp) = C

9)
For projectile motion, the vertical acceleration is the same throughout and is equal to g. There is no component of acceleration in the horizontal direction, so horizontal velocity is unchanged. But vertical speed changes due to which the resultant velocity changes.

15)
L is moved further to the right, so there is a bigger clockwise moment, meaning there need to be a bigger anticlockwise moment to counter it.
note the system is in equilibrium so there must be no net force and no net torque
immediately eliminate A and B because there is a horizontal H force that is unbalanced
now the logic comes for C, D.
for D, the reaction is shifted closer to the right (you can consider centre of mass shifted closer to right), so clockwise moment decreases, anticlockwise moment increases. Thus balances the system. so D is correct

24)
http://www.s-cool.co.uk/a-level/phy...finitions-of-stress-strain-and-youngs-modulus
We know that E is inversely proportional to the area, and directly proportional to the length.
Now, when we increase the length to 3L, the extension also increases 3 times → 3x.
But becoz we increase the diameter to 2D, we have increased the area by 4 times. A = π(d/2)^2 → A is now 4A.
And since extension is INVERSELY proportional to area, the extension now falls → x/4

So, the new wire's extension = 3x/4
Ans = B

30)
Lambda = 574nm just calculator's mistake

31)
What you don't get in this ?

33)
Q = It
Q = 0.01 * 1800 = 18C

34)

 

[Thought blocker]

how do you do these questions ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

questions 7, 14, 15, 16, 27, 30, 31, 36, and 37

7)
Area of the v-t graph is displacement so
10 above and -6.25 below..
add them = 3.75m

14)
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2

15)

16)
Power = Force x velocity
Force = weight = 80 x 9.81=784.8
Power = 784.8 x0.5 = 392.4W = 0.392kW

27)
find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D

30)
F=qV/d
F=(1.6x10^-19 x 12 x10^3)/25 x10^-3
F=7.7 x 10^-14 N

31)
10 A pass each through each cross section area as given
So 10 A means 10 C per second. So to find electrons.
no of electrons=10 C/1.6 x10^-19
n=6.3x10^19 electrons

36)
What you dont get in this ?

37)
he want the ratio of the V1/V2 so we need the distance of R1 from x divided by the distance of R2 from x...the distance from R1 from X is x and the distance of R2 from X is the total wire length - the distace of R1 from X so it is L-x so the answer is D x/L-x

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf
Can i get help with question 29 and 35
Thank you

29)
Use dsinθ = nλ
from first info find d
Then again use that formula and find θ when n is 2
After that subtract it from 15.4 degree you'll get C as the answer
35)

 

[swdj24]

I will come tomorrow, gotta sleep ._.

hey can you help me? asap thankssssssomuch

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf #7,9,20,23,29,32,33,36

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf #15,16,19,29,31,34

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf Q11,12,18,19,21,31,32,35 PLEASEEEPLEASSEE

 

[Thought blocker]

hey can you help me? asap thankssssssomuch

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf #7,9,20,23,29,32,33,36

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf #15,16,19,29,31,34

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf Q11,12,18,19,21,31,32,35 PLEASEEEPLEASSEE

Later... not now :/

 

[Ahmed H. Al-Neel]

question 37 from http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
question 12 from http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
question 33 from http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
question 9 from http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

 

[Thought blocker]

question 37 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf
question 12 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
question 33 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
question 9 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Later

 

[amal sharkawi]

 

[Rockstar RK]

Lemme solve ur doubts

question 37 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf
question 12 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
question 33 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
question 9 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

 

[Rockstar RK]

question 37 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf
question 12 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
question 33 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
question 9 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Q-37 is ans. A?

 

[Ahmed H. Al-Neel]

Q-37 is ans. A?

yeah its A but how come? i chose C which is wrong

 

[Thought blocker]

Lemme solve ur doubts

37)
The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).

In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.

12)
(1000)(5) - 10u = zero
so
10 u = 5000
u = 500 m/s

33)
R = ρL/A Yes ?
We know increasing extension, there is increase in length and decrease in cross sec are..
so here L increases so R will increase

9)
First find the time, using S = Ut + 0.5t^(2) ; S = 1.25 , U = zero, t = ?
That is 1.25= 0.5 x 9.81 x t^(2), t = 0.5 seconds.
Now for the Velocity, we know it equals Displacement / time taken ----> V = 10 / 0.5 = 20 m/s so in this type of question either of 2 quantities are given, we just need to find 3rd and input them all in second eqn to get the final result we are asked.

 

[Rockstar RK]

37)
The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).

In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.

12)
(1000)(5) - 10u = zero
so
10 u = 5000
u = 500 m/s

33)
R = ρL/A Yes ?
We know increasing extension, there is increase in length and decrease in cross sec are..
so here L increases so R will increase

9)
First find the time, using S = Ut + 0.5t^(2) ; S = 1.25 , U = zero, t = ?
That is 1.25= 0.5 x 9.81 x t^(2), t = 0.5 seconds.
Now for the Velocity, we know it equals Displacement / time taken ----> V = 10 / 0.5 = 20 m/s so in this type of question either of 2 quantities are given, we just need to find 3rd and input them all in second eqn to get the final result we are asked.

I was about to post the answers You made it easy
Also u have done a mistake in question 2.. Though the answer remain the same due to the zero u have equated!!!!
It had to be: -5000+10v=0 because in question it is given canon recoil speed is 5m/s...

 

[Thought blocker]

A

I was about to post the answers You made it easy

Anytime*

 

[Thought blocker]

I was about to post the answers You made it easy

So will you help me... ??

hey can you help me? asap thankssssssomuch

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf #7,9,20,23,29,32,33,36

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf #15,16,19,29,31,34

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf Q11,12,18,19,21,31,32,35 PLEASEEEPLEASSEE

I'll solve s13_13/12
and you solve s13_11
Okay ?

 

[Rockstar RK]

So will you help me... ??

I'll solve s13_13/12
and you solve s13_11
Okay ?

Clarify it plsss, label the question and answers too

 

[Student12]

Question 3, 4 , 5 , 8 , 11 , 14 , 15 , 19 , 24 , 33 , 35 , 37 , 38 PLEASEEEE!http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf

Yaaar guys need help

 

[Thought blocker]

Clarify it plsss, label the question and answers too

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf Q11,12,18,19,21,31,32,35 PLEASEEEPLEASSEE
Solve this.. I am doing other two links...

 

[chymera]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf




question 6

any one please