Physics: Post your doubts here!

[Mubasher96]

^ I still dont get this part

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.

 

[Sana101]

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 , 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.

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?

 

[Sana101]

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

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

 

[Sana101]

PAPER 1 OCT/NOV 2010 /11
I need help with Q)13,18,19,24,33,34
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_11.pdf

 

[Sana101]

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]

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!

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

 

[confused123]

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?

yeah i guess so but i am not sure

 

[confused123]

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

No idea dre must be some area of the loop formula or we might have to use integration.....I HOPE SOMEONE ELSE ANSWERS THIS! THANKS

 

[confused123]

PAPER 1 OCT/NOV 2010 /11
I need help with Q)13,18,19,24,33,34
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_11.pdf

13)clockwise moment = 15 into 3 = 45 Nm , and anticlockwise 5 into 2 plus 10 into 2 , )the 10 N force is acting anticlockwise u can imagine by thinking the lever as your arm and you will get to know that the 10N force is actually giving a anti clockwise moment) so the difference is 45-30 = 15Nm i.e A

19) Pressure = p g h , 1020 into 9.8 into 10,000 = watever the answer u have to add the atmospheric pressure as well which will be 10^5. ans is D

 

[ABDSyed]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_23.pdf
CAn Any help me with the [Q 1(a)i]

 

[MEGUSTA_xD]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_23.pdf
CAn Any help me with the [Q 1(a)i]

A change of one percent. Therefore 2.05X(1/100)=0.0205
But, 2.05+(or minus)0.0205= 2.0705, so this is not 3sf. So we need to take 2.05X(1/100)=0.02

 

[khanrocks20]

when a charged particle enters perpendicularly in a uniform magnetic field it follows a circular deflection. the direction of force is obtained by using flemings left hand rule. CAN U PLZ EXPLAIN THIS SENTENCE?

 

[Scafalon40]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s09_qp_02.pdf
Q.2 part b(i)
isn't the change in momentum suppose to be mv-mu?
I can't solve this, please help

 

[confused123]

when a charged particle enters perpendicularly in a uniform magnetic field it follows a circular deflection. the direction of force is obtained by using flemings left hand rule. CAN U PLZ EXPLAIN THIS SENTENCE?

if u can bring any past ppr question regarding this then i might be able to help.

 

[étudiante]

when a charged particle enters perpendicularly in a uniform magnetic field it follows a circular deflection. the direction of force is obtained by using flemings left hand rule. CAN U PLZ EXPLAIN THIS SENTENCE?

flemings left hand rule..... drctn of magnetic field is first finger, drctn of movement of particle is seond finger (if its an electron den pint finger OPPOSITE to direction of movement....)
the thumb will give u drctn of da centripetal force dat causes da particle 2 move in a circular path....

 

[umarashraf]

flemings left hand rule..... drctn of magnetic field is first finger, drctn of movement of particle is seond finger (if its an electron den pint finger OPPOSITE to direction of movement....)
the thumb will give u drctn of da centripetal force dat causes da particle 2 move in a circular path....

object moving in a circular path experience a so called centripetal force.... they why objects leave the circular path making a tangent to its position of release...??? rather it should fall toward the centre...????

 

[ABDSyed]

A change of one percent. Therefore 2.05X(1/100)=0.0205
But, 2.05+(or minus)0.0205= 2.0705, so this is not 3sf. So we need to take 2.05X(1/100)=0.02

Thanks Alot Bro

 

[Mustehssun Iqbal]

object moving in a circular path experience a so called centripetal force.... they why objects leave the circular path making a tangent to its position of release...??? rather it should fall toward the centre...????

Good question! Considering the Newton's second law, F= ma, the direction of acceleration is towards the direction of the applied force. Rather the object moves in a circular path.
There is also a condition for an object to move in a circular motion. That is, the object is projected with an instantaneous velocity in a certain direction. Suppose there's a car moving with an instantaneous velocity of say, 20 metres per second. Horizontally, it's moving with a constant speed- it's in motion. Vertically, it's acceleration is zero because mg is equal to the normal reaction force R. Then the car , suppose jumps over a speed breaker and now the component of it's horizontal velocity is still the same as it was before, but now the net force applied vertically on the car is not zero. Now it has an acceleration downwards, and hence it also has a speed, which is not uniform downwards. So now the car has two components of speed. And the car follows a kind of a circular path.
When a satellite experiences the gravitational force, you might have studied it in gravitation , that the satellite moves in a circular path due to the centripetal force of earth on it. That centripetal force is constituted by weight of the satellite, in other words, by gravitational pull. In that case, the satellite would have moved in the direction towards earth and fall somewhere on it. But that doesn't happen often. In that case, the satellite is also launched with a certain velocity so that it moves in a parabolic, or you can call it circular, or a rotational path, instead of following a tangential path.
Conversely, The satellite experiences , or has two components of velocity , and the resultant motion due to those components, you could say, is a parabolic path too.
It also relates to projectile motion. What happened in projectile motion was, there are two components of an object 's velocity, one horizontal and the other vertical. Horizontal component was constant most of the time, but horizontal component of the falling object changed. There was uniform acceleration on the object and there was gravitational force on the object. The object moved in a parabolic, or a circular path as a result of the two components of velocity on it, one almost continuously changing in magnitude.
And there's a very good example of rotating a water bucket vertically. That comes in the part of the topic, vertical motion, though it can also be used to explain this phenomenon. Suppose you are rotating a bucket of water , tied with a rope, constantly in vertical motion. And crazy it might sound, the bucket of water is uncovered. : D
When the bucket of water is on top of the circle, and is pointing downwards, the water would be observed to fall. But the water doesn't fall in that case, when you launch it into a circular motion with speed very high.
Actually, the water in the bucket is undergoing circular motion too. So when the instantaneous velocity of the water is very high in magnitude, the water still experiences a downward acceleration, but it follows a circular path too. That way the water doesn't fall from the bucket in circular motion too.
I've heard lectures on circular motion from very good teachers, and this one is also awesome.
If you aren't too busy, you can also take a look at this MIT lecture. I know it's MIT , but it's still good to clear many concepts regarding circular motion. : )
http://ocw.mit.edu/courses/physics/...mechanics-fall-1999/video-lectures/lecture-5/

 

[Mustehssun Iqbal]

object moving in a circular path experience a so called centripetal force.... they why objects leave the circular path making a tangent to its position of release...??? rather it should fall toward the centre...????

Apologies I misread your question!
The object moves in a tangential path because there's no more centripetal force acting on the object. Centripetal force on the object acts towards the centre of the circle, and when it is removed, there's usually only one component of velocity remained of the moving object, and that 's the component which we refer to as the instantaneous velocity of the object in circular motion at any point in the circle. Since there's just one component of velocity on the object, it then follows a tangential path rather than a circular one. It follows a circular path because it had two components of velocity, one constant and one changing continuously. When the centripetal force is removed - when one of the components of motion ( generally) is removed, there just remains one component of velocity on the object, and that is tangential.
Suppose you rotate a bob attached with a thread in a circular motion. During the motion, you cut the thread with scissors. The object similarly doesn't follow a circular path, but moves in a tangential path too.

 

[Mustehssun Iqbal]

object moving in a circular path experience a so called centripetal force.... they why objects leave the circular path making a tangent to its position of release...??? rather it should fall toward the centre...????