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help pls im stuck in this......T1 sin(50) + T2sin(40)=7.5... how to continue the equation to get T1 nd T2, pls i need a quick reply with details..thanx very much
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and the 2nd?acceleration should be constant...
i got it logically but i cant do the working..plz help meone individual spring will have an extension of x... when connected in parallel they will have an extension of .5x.... extension will be divided equally between the 2 springs... when another spring will be connected in series to them, the extension will again be added up... .5x+x giving 1.5x...
see in parallel springs we calcualte spring constant by applying this formula: k + k ...yes plz explain..
Jazakallah i get it nowsee in parallel springs we calcualte spring constant by applying this formula: k + k ...
and springs in series the spring constant is calculate by 1/k + 1/k....
so there are two springs in paralel F= kx , spring constant for these two springs is k + k = 2k
extension for these two springs is 2x
for the spring in series with these two springs the constant k is 1/k
extension for this spring is 1/2x
total extension:-
1/2x + 2x = 3x/2
Subhanallah i am not sure how did i calculated this answer. caz F = kx , i just added up kx from both springs...Jazakallah i get it now
yeah take several readings and then take the average, avoid carelessness while taking readings, reducing human errorguyz can u tell me 3 ways of reducing random errors other than "repeating the experiment"
what i got is that in order to have negligible resistance the weight should be more then the upward force, and it is after calculation. if drag forces were present then the forces would be closer or equal. not the case here.9702 s06 qp 2
Q1 c part ii
in the ms 2 justifications are provided...can sm1 elaborate dem plz
Please answer the question above!Please help me with this question In physics paper 23, question 2 part (c). I have no idea about vector triangles. http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_23.pdf
Thank you very much.
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 furtherwhat i got is that in order to have negligible resistance the weight should be more then the upward force, and it is after calculation. if drag forces were present then the forces would be closer or equal. not the case here.
i can't. you need to discuss with some teacher.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
6.i)draw arrows in the direction of the field. i.e downwards for the positive charge and upwards for the negative one.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!
Why is the direction the way it is? I don't get it.6.i)draw arrows in the direction of the field. i.e downwards for the positive charge and upwards for the negative one.
i was expecting a thanks for the other answers but anyway..Why is the direction the way it is? I don't get it.
I - uh - didn't ask for the other answers.i was expecting a thanks for the other answers but anyway..
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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