Physics: Post your doubts here!

[littlecloud11]

Initial velocity is 8.4, so your starting point is (0, 8.4). The time taken for the ball to reach the ground is .47 second and the velocity with which it reaches the ground is 12.9, so your second point is (.47, 12.9). The velocity increased and the ball accelerated so it's a straight line with an increasing gradient. The ball remained in contact with the ground for .02 s so there was a time lapse within which the velocity changed, this is shown by the dotted line. The velocity fell from 12.9 to -4.2 as the direction is now reversed and the ball begins to move up.so your next point is (.49, -4.2)
the rebound velocity is 4.2. Use v = u+ at to calculate the time after which the ball comes to rest. 0 = 4.2 + 10* t, t= .42 sec
total time the ball traveled =.47 + .02+ .42 = .91 sec and it's final velocity is 0 m/s
so your final point is (.91, 0)

salvatore

 

[Yousif Mukkhtar]

Can some one explain Q 3 b iii)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf

 

[hellangel1]

What would be the graph?

 

[salvatore]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf
Please help me with qn no. 6(b)(ii).. I'm totally confused!
Thans

 

[gary221]

What would be the graph?

which ppr??

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf

Q6 (b) (i) State and explain the polarity of the pole P of the magnet.

How do you figure this out? :s

Yup..i was confused too..
Bt see, the weight on the balance increases, which means there is a force on the magnet acting DOWNWARDS.
According to Newton's 3rd law (action-reaction law), there is an UPWARD force on the wire..
And so, according to Fleming's left hand rule, P is the north pole.
Hope u gt it!!

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
Please help me with qn no. 6(b)(ii).. I'm totally confused!
Thans

The phase diff is 180 as only when the two waves cancel each other out will a dark fringe be formed -----> http://www.s-cool.co.uk/a-level/phy...it/diffraction-interference-and-superposition

 

[salvatore]

The phase diff is 180 as only when the two waves cancel each other out will a dark fringe be formed -----> http://www.s-cool.co.uk/a-level/phy...it/diffraction-interference-and-superposition

Thanks..

And what about part 2, the ratio question?

 

[gary221]

Thanks..

And what about part 2, the ratio question?

On it...
okay, so max (light fringe) amp = add the 2 amps (A + B) = 2+1.4 = 3.4 units
min amp(dark fringe) = A - B = 2-1.4 = 0.6 units
n since intensity is proportional to amplitude^2
thus, (3.4)^2 / (0.6)^2 = 32
Hope u gt it!!

 

[snowbrood]

On it...
okay, so max (light fringe) amp = add the 2 amps (A + B) = 2+1.4 = 3.4 units
min amp(dark fringe) = A - B = 2-1.4 = 0.6 units
n since intensity is proportional to amplitude^2
thus, (3.4)^2 / (0.6)^2 = 32
Hope u gt it!!

can u answer q .5 b(ii) from physics coursebook exam style questions page no 118

 

[gary221]

can u answer q .5 b(ii) from physics coursebook exam style questions page no 118

which bk??

 

[snowbrood]

which bk??

cambridge internation as and a level coursebook from david graham jones richard and chadha

 

[snowbrood]

which bk??

could u just confirm answers i got density as 666.7 and pressure at L as 4578

 

[salvatore]

On it...
okay, so max (light fringe) amp = add the 2 amps (A + B) = 2+1.4 = 3.4 units
min amp(dark fringe) = A - B = 2-1.4 = 0.6 units
n since intensity is proportional to amplitude^2
thus, (3.4)^2 / (0.6)^2 = 32
Hope u gt it!!

Oh yeah! Thanks

 

[salvatore]

can u answer q .5 b(ii) from physics coursebook exam style questions page no 118

I just interfered!
For b(i) 800 × 0.5 × 9.81 = ρX × 0.6 × 9.81 [1]
ρX = 666 ≈ 670 kg m−3
= 666 kg/m^3

(iv) ρhg = 666 × 9.81 × 0.7 [1]
pressure = 4580Pa

Your answers are correct!

 

[snowbrood]

I just interfered!
For b(i) 800 × 0.5 × 9.81 = ρX × 0.6 × 9.81 [1]
ρX = 666 ≈ 670 kg m−3
= 666 kg/m^3

(iv) ρhg = 666 × 9.81 × 0.7 [1]
pressure = 4580Pa

Your answers are correct!

u just helped thanks man i just wanted answers thanks anyways

 

[gary221]

salvatore, thnx!!

 

[soul]

can someone please explain to me Q 1 ii) and c) and Q7
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf

 

[xxfarhaxx]

must watch!!

 

[gary221]

What would be the graph?

since g is inversely proportional to r^2
when r = R, g = 1.oo g
r = 2R, g = 0.25g
r = 3R, g = 0.11g
r = 4R, g = 0.0625g
plot the points n voila....ur graph is made!!