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Physics: Post your doubts here!

confused123

confused123

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Sana101 said:
PAPER 1!
Can someone please explain me the following Qs) 2,13,14,15,18,21,25,29
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s09_qp_1.pdf
Click to expand...

2) in this mcq you are given with the angle and horizontal component.use it to find the vertical comp.
Tan30 = Perpendicular / 20 , it comes 11.5N i.e C.

13) no convincing ans but the weight is making a torque of 180 Nm, how? 900 N into .20m (.20 is the perpendicular distance or the radius of the disc)
same torque will be shifted upwards in order to lift this weight i.e 180 = F into 1.20 m it comes 150 N hence B. ( i dnt knw why i took .20 as the perp distance to find the torque created by weight)

14) Power = F v , so k = P over v^3 because you need to add another v in the denominator as F = Power over velocity.

15) what i get is dat if u c in both vessels the height will decrease by half from the original height in the vessel X, means h will bcm h/4 altogether, so the answer will be B mgh/4.

18) pressure = pgh , p ang remain constant. u know that h has increased in the right side of manometer. i.e pressure has increased. there is only option D 2pgh which supports this idea. other three options shows decrease or no change in the pressure.

21) strain energy = 1/2 F into x so 1/2 into 17 into 30 /100 it comes around 2.5 J but the best estimate would be less then that i.e 2 Joules which is nearest to the whole number. 3J would be very higher i.e A option

25)i dnt understand :mad: , it shud be C but its A, d sin Q = n lambda , this s what comes after moving here nd dre things
n/N = sin Q / Lambda
29.
 
confused123

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Sana101 said:
PAPER 1 m/j 2010
I need help with the following questions please - 5,9,15,16,20,29
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_11.pdf
Click to expand...

5 is c, test it urself take force to be 10N and increase the vertical component angle sin theta i mean, then do it same for the horizontal cos component.


9)B it is displacement time graph whose gradient will show us the velocity. velocity is gonna increase every second aat the rate of 9.8m /s/s . so B shows a graph like dat..

15)pretty simple. use the formula Power = F into v
total force included force of the load and the crane. load's mass is 1000kg , apply W=mg to find the force....this will be 11000 into .50 = 5.5kW hence B

20) Force= spring constant into extension or compression. take any force from the graph say 8 N see the extension its 60 mm minus the original spring legth which is 40 mm extension comes 20mm ! convert 20 mm into meters. it comes 200N/m at the end..
 
XPFMember

XPFMember

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Mubasher96 said:
now this is the minimum area the rod should have, so that it does not break.(lesser shall it be, the rod would break.) and this can only be such when the area of bubble is its maximum.
(max) area of cross-section = (3.2 – 2.0) × 10–6
= 1.2 × 10^–6
when bubble has 1.2 x 10^-6, rod has 2 x 10^-6, a total of 3.2 x 10^-6
Click to expand...
^ I still dont get this part :(
 
T

Tina

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Guyz, i need help in paper may/june 2005: Q2/(a);Q5/(b);Q7/(a).

Please post answer with graphs. Thanks!!
 
Mubasher96

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XPFMember said:
^ I still dont get this part :(
Click to expand...
the bubble reduces the area of the rod. for rod not to break, the limit is to have an area of atleast 2 x 10^-6
iss se kam hua to it'd break, that means if bubble has more area than its max., the rod wud break as a result of decreased area (rod's). the total is 3.2 etc which *should* have atleast 2 x10^-6 for rod and the rest for bubble.
hope it helps. :)
 
S

Sana101

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confused123 said:
2) in this mcq you are given with the angle and horizontal component.use it to find the vertical comp.
Tan30 = Perpendicular / 20 , it comes 11.5N i.e C.

13) no convincing ans but the weight is making a torque of 180 Nm, how? 900 N into .20m (.20 is the perpendicular distance or the radius of the disc)
same torque will be shifted upwards in order to lift this weight i.e 180 = F into 1.20 m it comes 150 N hence B. ( i dnt knw why i took .20 as the perp distance to find the torque created by weight)

14) Power = F v , so k = P over v^3 because you need to add another v in the denominator as F = Power over velocity.

15) what i get is dat if u c in both vessels the height will decrease by half from the original height in the vessel X, means h will bcm h/4 altogether, so the answer will be B mgh/4.

18) pressure = pgh , p ang remain constant. u know that h has increased in the right side of manometer. i.e pressure has increased. there is only option D 2pgh which supports this idea. other three options shows decrease or no change in the pressure.

21) strain energy = 1/2 F into x so 1/2 into 17 into 30 /100 it comes around 2.5 J but the best estimate would be less then that i.e 2 Joules which is nearest to the whole number. 3J would be very higher i.e A option

25)i dnt understand :mad: , it shud be C but its A, d sin Q = n lambda , this s what comes after moving here nd dre things
n/N = sin Q / Lambda
29.
Click to expand...


Thank you so much! this really helped! just a little doubt here...for question 21) nearest estimate from 2.55 would be be 2.6. why do we pick 2.0 then? is it that the answer for these type of questions should always be a whole number?
 
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Sana101

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confused123 said:
i small square represents 1 mm so multiply with the number of squares. or dre maybe some formula to calculate the area of the loop
Click to expand...

if for example we take this measurement in cm then
1 small square = 0.1 cm
1550 small squares= 1550x0.1= 155 cm
converting this to m will give 1.55m! the answer in the ms is 0.031m?! 0.O
 
S

Sana101

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PAPER 1 OCT/NV 2010 /12
I need help with Q) 4,8,11,14,21,22,26
Q40) why isn't the answer 'A' . The half life IS achieved after 90 min!
 
umarashraf

umarashraf

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Sana101 said:
PAPER 1 OCT/NV 2010 /12
I need help with Q) 4,8,11,14,21,22,26
Q40) why isn't the answer 'A' . The half life IS achieved after 90 min!
Click to expand...

i think there is nothing as such thing mentioned that the counter has no background radiations.... so we can necer be certain....
 
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