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Physics: Post your doubts here!

confused123

confused123

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anonymous123 said:
i didnt get the first statement...in almost every case the weight of a body is greater thn resistance for some tym b4 it reaches its terminal velocity. i do get that the resistance is very small compared to the weight in dis case but wht if the resistance was 0.01N cud it still be neglected? dat is smaller thn the weight so can we ignore it as the ms seems to imply? i dnt think so..plz explain further
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i can't. you need to discuss with some teacher.
 
confused123

confused123

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Jaf said:
What is the direction of force on the charges in this question and why? (6b) :
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf

The mark scheme doesn't mention anything. The examiner report is unhelpful too. :|
Also, this is a question that has been repeated in a recent paper.

Thanks!
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6.i)draw arrows in the direction of the field. i.e downwards for the positive charge and upwards for the negative one.

ii)we are given with the electric field strength in the a part which is : 2 into 10^3 V/m

Electric field strength = Force / Q

2 into 10^3 into 1.2 into 10^-15 = Force , force is 2.4 into 10^-12 N

iii) Torque of a couple = Force into perpendicular distance between two forces

2.4 into 10^12 into 0.0025 sin 35 ---> torque is 3.44 into 10^15 Nm

iv) the subsequent motion is to rotate in order to align with the field.

P.S these are basic concepts and formulas maybe that's why marking scheme and examiner reports didn't go into details.
 
confused123

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Jaf said:
Why is the direction the way it is? I don't get it.
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i was expecting a thanks for the other answers but anyway..:cautious:

i guess the arrow with the positive charge will be towards right and down and the arrow with the negative charge will be towards left and up together trying to rotate the object
 
Jaf

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confused123 said:
i was expecting a thanks for the other answers but anyway..:cautious:

i guess the arrow with the positive charge will be towards right and down and the arrow with the negative charge will be towards left and up together trying to rotate the object
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I - uh - didn't ask for the other answers. :unsure:

And are you sure you meant to say right AND down, and left AND up instead of right and left OR down and up? Because the former wouldn't make sense. You'd have directions pointing in both, the clockwise and anticlockwise directions. :/

Part of my original question remains unanswered. Why would they behave in a such a way? Are charges supposed to line up parallel to the direction of the field?
 
B

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What is the effect of springs in series and parallel on extension and spring constant? can somebody plz help me with this...

+

does anybody know which derivations are in our course?
 
confused123

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Jaf said:
I - uh - didn't ask for the other answers. :unsure:

And are you sure you meant to say right AND down, and left AND up instead of right and left OR down and up? Because the former wouldn't make sense. You'd have directions pointing in both, the clockwise and anticlockwise directions. :/

Part of my original question remains unanswered. Why would they behave in a such a way? Are charges supposed to line up parallel to the direction of the field?
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well u asked for the 6 b part. :|

i am pretty sure now that the direction of force on the positive charge is towards right parallel with field line, and on the negative charge its towards left because according to my concept that's how a couple of a force is formed. the forces are equal and opposite in direction. the positive one will tend to move right as it get attracted to the negative plate and the negative charge will be attracted towards positive plate i.e on left. what say?
 
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hmlahori

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confused123 said:
q.2 b) if u will suspend a plumbline from different positions then you can know the centre of gravity. i mean you need to draw 4 lines from vertices and then find their intersection point, that will be below pivot. it will eventually come to rest because centre of gravity is in line with pivot. no turning effect produce as we are applying force on the pivot line. this is what i got from the question.

4 b ii) lets find the approximate are:2 [1/2 into .3 into 0.08 ] = 0.024 m

note: i converted the speed from cm/s into m/s and divided the graph into two triangles.

s 06) q.2 c) resolving forces horizontally: we can see that F cos a and T cos B are the only two horizontal forces. now just equate them, they must be equal so that the system remains in equilibrium. Fcos a = T cos b

ii ) resolving vertically: there are two forces acting upwards which are F sin a and T sin b , these two upward forces must equal with the downward weight. hence Fsin a + T sin b = W

iii) sum of moments at A) F doesn't contribute as it is acting at the pivot. o moment by force F.
now we know that AC = 2/3 Ab
the upward force is T sin B and downward is W. the distance of W from A in terms of AB is 2/3 , the distance of T sin B is equal to AB from A point. . now the equation becomes 2/3 AB into W = T sin b into AB , which further becomes 2W= 3T sin b. hope that helps.

can you give your opinion on this question b part?
Click to expand...

question 4 b basically asks how rod will be in equilibrium. For this to happen the resultant force on the rod must be zero as well as any moments of forces. the two other forces W & T pass through P and so force F must also pass through P. in this way the distance of all the forces together from P will be zero hence they will cancel out the effect of each other and no resultant force or moment about P thus rod in equilibrium.
And thanks for your help!
 
confused123

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hmlahori said:
question 4 b basically asks how rod will be in equilibrium. For this to happen the resultant force on the rod must be zero as well as any moments of forces. the two other forces W & T pass through P and so force F must also pass through P. in this way the distance of all the forces together from P will be zero hence they will cancel out the effect of each other and no resultant force or moment about P thus rod in equilibrium.
And thanks for your help!
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that's nice. thankyou :) , how will be the distance from P together of all forces be zero, how will they cancel i mean :/
 
Jaf

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confused123 said:
well u asked for the 6 b part. :|

i am pretty sure now that the direction of force on the positive charge is towards right parallel with field line, and on the negative charge its towards left because according to my concept that's how a couple of a force is formed. the forces are equal and opposite in direction. the positive one will tend to move right as it get attracted to the negative plate and the negative charge will be attracted towards positive plate i.e on left. what say?
Click to expand...
Oh wow. Yes, this is it! Thanks so much! :D
 
dilchan

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confused123 said:
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s08_qp_2.pdf
q.6 b
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Switch closed means the power pass through that part of circuit(through the closed switch) and as the heating elements are similar they all have the same constant resistance. So calculate the parallel resistances and thereby calculate the TOTAL RESISTANCE of the heating elements in circuit where the current passes. The calculate the power using the total resistance. The power supply is providing a constant emf of 240v. Contact me for more details.
 
sidbloom

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i need to know what is exactly meant by shorted lamp..........
by the name i think it means that the lamp is fussed but when i searched it i got really confused....so i want to know what is the exact meaning
thanks ^_^
 
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