Physics: Post your doubts here!

[Thought blocker]

Q10) This one has been discussed before, here are a few posts on that:

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-318#post-692472

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-567#post-828639

Q21) I'm pressed for time right now, i'll try posting it later.

Good Luck for all your exams!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_13.pdf

Q21 (B) Q10 (A)
Urgent!! Im sorry

ZaqZainab sagar65265

21)
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

 

[Zepudee]

21)
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

Thanks thought blocker and sagar! All the best

 

[Thought blocker]

May be my last post ova here all the best and good bye….…

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
Q15

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
Q15

unique111 posted another solution here (that is totally correct, despite that last line!) - I'd posted at around the same time, so the following is a straight "lift" off that comment, in case you'd like a little elaboration.

When a total amount of work W is done on any object, the kinetic energy of that object changes by a quantity W - if negative work is done on the object, the kinetic energy changes by a negative amount and if positive work is done on the object it's energy changes by a positive amount.

Suppose you throw an object upwards, gravity is the only force that does any work on the object. That work is negative, so the kinetic energy of the object decreases until the object comes to a stop. When it falls down, gravity does positive work on the system thus increasing it's kinetic energy.

In this case, taking the first situation, the force is constant at magnitude F, over the entire displacement s. Thus, the work done by that force is Fs. By the
Work-Kinetic Energy theorem, this is equal to the increase in kinetic energy of the system. This value is given as 4 Joules (8-4 = change in KE = +4 Joules).

In the upcoming situation, the force is 2F, displaced through a distance 2s. Thus, the total work done is 4Fs. This is also the change in kinetic energy of the object.
From above, we know that Fs is 4 Joules, so 4Fs = 4*4 = 16 Joules increase. From 4 Joules, the increase of 16 Joules takes it to 20 Joules = B.

Good Luck for all your exams!

 

[..sacrifice4Revenge..]

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
14

 

[maq1996]

worst paper ever

 

[kitkat <3 :P]

worst paper ever

agreed

 

[maq1996]

im just so disappointed.. dont think id be getting more than 23

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
14

This is from a post that is two pages back, iin case you have any more doubts from this paper, you may find the answers there and on the page after that (p. 578 and p. 579):

Q14) When the projectile reaches it's highest point, it has a purely horizontal velocity and no vertical component of velocity.
Additionally, since there is no force on the projectile in the horizontal direction, it has the same horizontal velocity at it's highest point as it did when it was shot.

A little more on this - suppose the initial velocity has a magnitude of "v", the initial kinetic energy E = 1/2 * m * v² = 0.5mv².

Now, onto the focus of the question: since the initial velocity points at an angle 45° to the horizontal, the component of velocity in the horizontal direction should be

v(horizontal) = v * cos(45) = v/(√2)

This is, by our discussion above, the velocity of the projectile at it's maximum height. So, the kinetic energy at it's greatest height is

E(f) = (1/2) * m * (v/(√2))² = 0.5m * v²/2 = 0.25mv².

Let's divide E(f) by E, giving us

E(f)/E = (0.25mv²)/(0.5mv²) = 1/2
Therefore, E(f) = (1/2)E = 0.5E = A.

Good Luck for all your papers!

 

[Student12]

It was difficult

 

[danial 234]

Time wasnt sufficient... made some random guesses ... overall it was fine.. neither very good.. nor very bad

 

[Wolfgangs]

Were there some repeated questions from pastpapers? At least 25%? Please mention it in terms of percent XP

 

[26697luj]

what varient did you do?

 

[26697luj]

guys just the varient its not like i am asking for the questions!

 

[maq1996]

what would the threshhold be around?

 

[..sacrifice4Revenge..]

what would the threshhold be around?

https://www.xtremepapers.com/community/threads/physics-12-gt.34503/#post-829148

 

[Rockstar RK]

Bro, I am an external student in my school.Today when I was solving the question paper, the Exam Officer and Supervisor repeatedly were disturbing me by sliding the OMR sheet between my vision and at times in between my pen's tip. This happened 5 to 6 times which caused loss of time. So what should I do to revert this act by the authorities.

 

[Anonymous981]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf
Q8