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q4, its C, the division is by 0.4
and 18, i think its just mgh? i 780J, post the answers
Q9 : D
Q18 : C
Q4 isnt coming right
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q4, its C, the division is by 0.4
and 18, i think its just mgh? i 780J, post the answers
Hmm... Alright.variance 3, so taking paper 13
Q4 : D
Q18 : C
Q4 isnt coming right
i dont know how its D.. even if you maximize error its still 3.6 smh
you resolve the motion into horizontal and vertical components
thanks man !!!!duuude, easy. full wave length from the diagram is six blocks, means 6x0.2ms, so 1.2ms, so 1.2x10^-3, but f=1/T f=1/1.2x10^-6, you will get 833.33 so about 830hz, is that the right answer?
thanks man !!!!duuude, easy. full wave length from the diagram is six blocks, means 6x0.2ms, so 1.2ms, so 1.2x10^-3, but f=1/T f=1/1.2x10^-6, you will get 833.33 so about 830hz, is that the right answer?
th
thanks man !!!!
alright the answer is D to that, listen up, the horzontal velocity always remains the same and is the one which causes the horizontal distance moved, so from velocity = distance/time, you can use that, to find time use the formula s=0.5gt^2, after transformation to make t subject you will get t=root(2s/g) plug in the t formula in the velocity formula, mind you s is the vertical distance, and in the velocity formula its the horzontal, you willl get the answer
you resolve the motion into horizontal and vertical components
the vertical initial velocity is 0 m/s with -9.81 m/s^2 acceleration and the distance is 1.25
so we apply the equation
s(displacement) = ut + (gt^2)/2
1.25 = 0 + (9.81 x t^2)/2 = 4.905 t^2
square root, so t= 0.505seconds
now we solve the horizontal component, since velocity is unchanged we can use the basic formula
speed = distance / time taken= 10/0.505=19.8 answer is D
i think its C .. by my own made up theories
F(net)= W = mg= (120 - 80) x 9.81 = 400
help needed
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w11_qp_11.pdf
16 (c)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s11_qp_11.pdf
4 (c)
14 (c)
16 (b)
25 (d)
35 (c)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_12.pdf
12 (c)
15 (d)
THANK U VERY MUCH
for w11
F(net)= W = mg= (120 - 80) x 9.81 = 400
Fnet = ma = (120 + 80) x a = 400 a=2
kinematics equation
V^2= u^2 + 2as
v^2 = 0 + 2x2x9
v = 6
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