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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
Can any one help me with the Question No:7,15,22...
Please anyone
Q7--> First of all the acceleration is negative because "g" which is acceleration due to gravity is negative and constant... The small pulse is as u know at the time the ball hits the ceiling... Imagine it like that: when it hits the ceiling and bounces back its acceleration increases...there is a simple concept to know that the answer is D, simply because the puls is approximately in the middle of the line... not towards the end as in C
Q5- x is 0.44. Add uncertainties. U get 0.03. (0.03/0.44)*100 = 7%
Q8- find speed in both places. speed between X and Y = 40/12 and speed between Y and Z is 40/6. Then acceleration = change in velocity/time. so find the change in velocity and divide it by the average of time i.e. (12+6)/2
Q14- find the hypotenuse force 1st. 200/sin30. that will give u 400N. Then 400*1.5 gives 600J. Add frictional force and u get 750J.
Q22- Answer is a because it needs to have high sensitivity for small masses and low for high masses. The spring having smaller k stretches more and larger k stretches less. smaller k should be in the rigid box so it stretches till a certain extent. Hence small masses will make it stretch but when a large mass is present the small k spring will b stopped by the rigid box while the larger k will stretch less and show lower sensitivity.
Q25 & 26- sorry i suck in this chap :/
Q33- answer is d as the current had to be 2 not 1.8. just add both resistance and divide voltage by it to fins current. for example for a resistance was 10. so 2+10=12. then V=IR so I=12/12=1
Q34- votage decreases linearly so its supposed to be b
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
I know I have a lot of doubts.. will be really grateful for help
Q5, 8, 14, 22, 25, 26, 33, and, 34
Please anyone..
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
Q25,27,35.
Thanks in advance..
Q 22 - 1st of all calculate mass* 6.02*10^23. this gives u the mass of all atoms in the crystal. then find the volume by dividing the mass u calculated by density. density=mass/volume. Then divide volume by 6.02*10^23. and find its cube root
Q25- Umm i guess extension will be the same as it obeys hookes law. Not sure though :/
Q26 - speed=2d/t. distance is muliplyd by 2 as its reflected so it travels the same distance twice. speed is that of light. so (2*150)/3*10^8 will give u 1 microsecond.
Q28- the two points are very apart by almost 1 full wave. Half wave apart is 180 degree but here the separation is greater so its 270.
Q35- sorry cant explain ! bad at this chap :/
Q38 - Just use a similar value for all resistors like 5ohm and calculate the combined resistance. If u use 5ohm then circuit 1 has resistance of 2ohm, 2 has 3 ohm. and 3 has 3.75 ohm
Q7 : horizontal component of velocity in projectile remains constant. and vertical component first decreses till max height (where it is zero) and increases to max for the other half of the journey
the answer is A. This is beacuse pressure=hdg. so h=p/dg. pressure will now be 10% of atmospheric pressure. so (10/100)of atmospheric pressure. this gives atmospheric P/10 hence p/10dg
thnku so much sister for da helpthe answer is A. This is beacuse pressure=hdg. so h=p/dg. pressure will now be 10% of atmospheric pressure. so (10/100)of atmospheric pressure. this gives atmospheric P/10 hence p/10dg
no probthnku so much sister for da help
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