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Physics: Post your doubts here!

N

NokiaN95638

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Idolfanatic95 said:
In Q3 b (iv) why do we use 4.1/sin30 to calculate distance?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_22.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_22.pdf
In Q6 c (i) how do we calculate the phase difference? and how come it's 0 rad. ?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_23.pdf
Click to expand...

S11 22;
We know that frictional force always acts along the plane. Never at an angle to it. Therefore the frictional force acts along the line AB. Since we know
Work done= Force x distance. The distance AB is found by using 4.1/sin30. Hope it helps

S11 23:
Since the wave is a stationary wave all particles between 2 nodes are in phase. Since X abd Y are in phase the phase difference is 0 (as they are between 2 nodes)

Hope it helps you
cheers
 

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Unicorn

Unicorn

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robotic94 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_23.pdf
Q1 d
Please explain me with the triangle method!
Click to expand...

Ok trig.png

as you can see from the diagram I rearranged the forces but didn't change the direction
the angle between T1 and T2 is 90 because T1 is at an angle of 50 to horizontal and T2 at 40 to horizontal
and the angle between 7.5N and t=T1 is 40 because it is vertically downwards so is perpindicular to horizontal

to find T1 and T2 use trignometery

hope it helped
 
Idolfanatic95

Idolfanatic95

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NokiaN95638 said:
S11 22;
We know that frictional force always acts along the plane. Never at an angle to it. Therefore the frictional force acts along the line AB. Since we know
Work done= Force x distance. The distance AB is found by using 4.1/sin30. Hope it helps

S11 23:
Since the wave is a stationary wave all particles between 2 nodes are in phase. Since X abd Y are in phase the phase difference is 0 (as they are between 2 nodes)

Hope it helps you
cheers
Click to expand...
Helped :)
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_ms_2.pdf
Can u help me wid Q 1 (b) Im not getting the uncertainty in R right . .
 
Unicorn

Unicorn

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Idolfanatic95 said:
Helped :)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_ms_2.pdf
Can u help me wid Q 1 (b) Im not getting the uncertainty in R right . .
Click to expand...


ok the rules for dealing with uncertainity are

when adding/subtracting 2 terms togther ADD ABSOULTE error
When multiplying/dividing 2 terms togther ADD PERCENTAGE error
when a term is to a power of something say x^2 then MULTIPLY PERCENTAGE error with power

in the case multiply it with 0.5
 
Unicorn

Unicorn

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Idolfanatic95 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_ms_2.pdf
Q2 (d) How do we do it?
Q4 (d) ii= Range in the ms in 41-48 degree. I'm getting 55 degree :-s
Click to expand...

for question 2 you have
accn = 2.5 x10^15
time = 2.4 x10^-9
intial speed= 5 x10^7

and u know that the electron is launched halfway between the plates so the distance between it and the plate is 1.5/2 =0.75

using equation of motion= s= ut +(at^2)/2 you get s to be 0.1272 which is much less than 0.75 so no it does not hit it

for question 4 ur diagram is probably wrong draw it again
 
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