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Physics paper 1 AS level doubts here!!!

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Q)8 u first find speed between points X and Y which is 3.33m/s and find speed between Y nd Z which 6.67m/s
then u find average of the time taken which is 12+6/2 which is 9 sec
then u find acceleration v-u/t so 6.67-3.33/9 = 0.37m/s^2

Q)14 u first find the force of the person to lift the box at an angle of 30 degrees so sin 30 x 200=100N AND THEN u find the lenght of the slope as work done is force x distance moved by the object so it will be 1.5/sin 30 =3m then W=100 x 3 - -150 x 3=750J

Q15)I'll check out

Q21)For these type of questions...i strongly suggest u to assume numerical values yourself..according to question..like if it says P has diameter twice than that of Q..than assume diameter of q=2cm...and therefore P=4cm..these assumptions will make it easy for u

Q)27)??? I'll check it out later
 
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5. B
Frequency: 3000 oscillations in a min, ie, 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01
24. C
peak position, particles at rest.
particles leading crest move up.
Hey help me as well.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf
q 7,9, 10, 18, 22
Q7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory

Q9)use the formula v^2 = u^2 + 2as
where
positive direction = direction of train
u = original velocity (speed of train)
v = final velocity (zero)
a = acceleration of train (in this case, acceleration is negative)
s = distance (from point where velocity = u to where velocity = v)

The problem uses x for distance, not s, so the equation we'll use is v^2 = u^2 + 2ax

because v = 0, we can write
u^2 + 2ax = 0, rearranging as:
x = -(u^2) / 2a . . . . . . . . Note - a has a negative value, thus making x positive
x = Ku^2 where K = -1 / (2a)
ie. x varies as the square of u
so if u increases by 20% (ie changes by a factor of 1.2), then x will change by a factor of (1.2)^2
which is a factor of 1.44
thus the minimum distance between yellow and red must now be 1.44x

Q10)g of P=1/10 of g of Q
so weight at planet Q is 1/10 * x =1 s0 x=10N and mass is 10/10 =1 kg

Q18)At what rate does the motor provide energy this part of the question is asking to find power and power=F x velocity so force is m1-m2 x g x v

Q22) for 3m of metal X the extension is 1.5 x 10^-3 so for 1 m = 1.5 x 10^-3/3=5 x 10^-4
for 1m of metal Y the extension is 1 x 10^-3 so for 3m =3 x 10 ^-3 m
add them and u get 3.5mm
 
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5. B
Frequency: 3000 oscillations in a min, ie, 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01
24. C
peak position, particles at rest.
particles leading crest move up.
Hey help me as well.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf
q 7,9, 10, 18, 22
Q.18 energy is equal to net force multiply by velocity. we wil consider net weight upward nd multiply by velocity. so D is correct
 
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Q16)u find the horizontal and vertical components and then use pythogras theorem to find F
so 10 sin 30=5N
10 cos 30=8.66025N
5^2 +8.66025^2=100 root of 100 =10N

how do we know if for the components we have to use 30 or 60?
 
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TSZ

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Hey, i made a thread regarding w09 questions, cn u please answer dem
 
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