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5)http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
Q5, 11, 22, 26 and 34. Please don't ignore this post at least XP
x is 0.44. Add uncertainties. U get 0.03. (0.03/0.44)*100 = 7%
11)
The tension in the rope needs to balance out the weight of the student acting downwards, so since the total tension needs to be opposite and equal to the weight, Let tension in rope on left side be T1 and the tension on rope on right side T2 and :. Total tension = T1sino + T2sino
add them up... W= 2Tsino, rearrange to get T= W/ 2sino. Answer: B
22)
because since k is low for the first spring, the box will prevent it from breaking by over extending. Hence A
26)
We know that distance between two minima is given by λ/2 (example, distance b/w T and S is λ/2)
We also know that Microwave travels with the speed of light c i.e is 3 * 10^8 m/s
Given distance b/w each minima is 15 mm so total distance from T to P would be 15 * 4 = 60 mm
Now distance b/w:
T and S = λ/2,
S and R = λ/2
R and Q = λ/2
Q and P = λ/2
and them you get 2λ
so 2λ = 60*10^-3
hence λ = 0.03 m
We know that c = f * λ
So f = c / λ
f = ( 3 * 10^8 ) / ( 0.03 )
f=10GHz
34)
voltage decreases linearly so its supposed to be B : as rho increases, there would be more steepness in graph