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Aoa
any idea what the threshold for for As p2 and p3 might be?
any idea what the threshold for for As p2 and p3 might be?
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i'm also stuck with this number. Please, do inbox me, if somebody replies.
thank you !sin theta = 40/50
theta = 53
F= 8Cos53 (Since, there are two components of 4N i multiplied it with 2, 4*2=8)
Oct09
Q12. since the btm belt is slacked, so tension is only at the upper belt, torque at Q = 3.0 N m,
so 3.0 = T x (100 x 10^-3)/2
T = 60N
Torque at P = T x (150 x 10^-3)/2 = 4.5
Q13. assuming it is projected with velocity v, so the initial KE, E= (mv^2)/2
at the highest point, the vertical component of the veolcity = 0, so its left with v cos 45,
so final KE = 1/2 x m x (v cos 45)^2 = 0.5E
Q14. By conservation of momentum, 2 x 2 = 1 x v, v =4
so the 1kg trolley moves 4ms-1 to the right,
calculating the total KE of the both trolleys, which is 1/2 x 2 x 2^2 + 1/2 x 1 x 4^2 = 12J
assuming no energy is lost, the energy stored in the spring is converted to KE gained by the trolleys, so its 12J
Oct11/12
Q23. the area under graph is the elastic PE, so we can c that the answer is obviously (i hope its obvious for u too) B. not much that i could explain here
everything is given ...taking moments ..clockwise moments wa+fh=2Wa..... and in the question the corner of the wall is considered as the pivot
question 7 is direct application of formulas
for vertical displacement s= ut + o.5 gt^2 and horizontal is S=ut
question 10 : since f and m are directly proportinal ...heavier mass need more force to move .. so , masses given m and 3 m ...if F=1 N then 3/4 the force goes to 3m and 1/4th to m ( well this is how i could explain it )
q 14 : you need to focus on what question asks for
vertical component of force = 10000-9000=1000N
question 17 : take average current ...since its reduced over a period of time ...at one instant the current is not constant
so 100+20/2 = 60 x 8 = 480mC
question 36: earthed plate is considered positive so electric field is directed upwards and E=V/d (dnt forget to convert the units)
question 40 : here is a reply from one of the members (Taci12) :
Question 40:According to syllabus, we must know that a proton has a charge +e (the basic/elementary charge)Now according to this question, protons themselves are made up of particles called quarks. There are different types of quarks, each one with its own charge.Whatever the combination of quarks, the total charge on a proton must be +e. Hence the sum of the charges of all the constituent quarks in a proton must be + e.So simply check the sum of charges of each option to see if it is +e.option A: 3 x (-e/3) = -eoption B: (2e/3) + (-e/3) = +(e/3)option C: (2e/3) + 2(-e/3) = ooption D: 2(2e/3) + (-e/3) = +(3e/3) = +eanswer is D
Graphs Graphs Graphs Graphs, STUPID GRAPHS!
itney saarey hain, confuse ker k rakh diya hae ;(what about them??
itney saarey hain, confuse ker k rakh diya hae ;(
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf
Can someone explain
q4 a ii) and for b)i) What is the direction of the force?
Thanks a lot.for Q4 a ii)
Ek = p^2/2m
1/2mv^2 = (mv)^2/2m (since, p=mv)
1/2mv^2= mv^2/2 (m and m cancels out)
1/2mv^2 = 1/2mv^2. (Hence, Proven)
for Q4 b i)
find change in momentum first,
delta p = m*delta v
p= 0.035*(4.5+3.5)
p=0.28.
F=change in momentum/change in time
F=0.28/0.14
F=2.0N.
Thanks a lot.
Can someone explain this question?
Thanks again brother.for minimum force to be applied the angle is always 90 degree.
and for second part,
Torque = Force*distance
Force=Torque/distance
F=130/0.45
F=290N
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