IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: krtcobain82 on June 02, 2011, 06:15:31 pm
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Guys I need explanations for the following questions.
1) may/june 07 Paper 1
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s07_qp_1.pdf
Question no 37 and 40.
2) may/june 09 Paper 1
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s09_qp_1.pdf
Question no 5, 7 .
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Guys I need explanations for the following questions.
1) may/june 07 Paper 1
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s07_qp_1.pdf
Question no 37 and 40.
2) may/june 09 Paper 1
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s09_qp_1.pdf
Question no 5, 7 .
Q37) This is a tricky one. Electrical resistance of P - R - pl/A
Let's assume the cross-sectional area of the resistor P is A
Thus, R = px/A
Resistance of Q
Now, the resistance is the combined resistance of all the wires
Say for instance, the resistance one of those wires is px/a (a < A, p, because they're of the same material)
So as the wires are in parallel, the combined resistance => 1/r = 1/r1 + 1/r2 + 1/r3 + ...
= a/px + a/px + a/px + ....
1/r = na/px
r = px/na
Now, As the volumes of the materials used is the same for both resistors, and their length is the same, the cross-sectional area of resistor P
should be the sum of all the cross-sectional areas of the wires in resistor Q
Meaning, na = A
thus r = px/A = Resistance of resistor P
Answer : C
40) The final velocity depends on the acceleration. The one with the least acceleration has the lowest final velocity (v^2 = 2as)
Acceleration, a = F/m
F= EQ
E is constant because V and d are constant.
a = QE/m
E is constant, thus, a is lowest, when the Q/m of the particle is the lowest. That is with C, Li, 3/7
Answer : C
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5) Best to use elimination in these type of sums. The final displacement can't be zero, thus, you are left with C or D
It can't be C because it doesn't show "constant terminal velocity"
Answer : D
7) Well, to be honest, i thought the answer was C. Well, i'm not sure of A, but that's the answer. The second law deals with force being
the rate of change of velocity times mass, and the first law is about inertia (unless acted upon external force one).
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For no. 7 I think its like this. According to first law, the body remains in its state or uniform motion or rest until acted by an external force. If we take this mathematically then F=ma. In absence of external force the equation becomes 0=ma>> or a=0. a=0 tells us that either the body is moving in constant speed or is not moving at all.
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For no. 7 I think its like this. According to first law, the body remains in its state or uniform motion or rest until acted by an external force. If we take this mathematically then F=ma. In absence of external force the equation becomes 0=ma>> or a=0. a=0 tells us that either the body is moving in constant speed or is not moving at all.
F=ma is the mathematical expression for the second law (Force is the rate of change of momentum). I don't think it's for the first.
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Yes, F=ma is the second law. For that question, no. 7, the solution as you said is option A (The first law follows from the second law.) and i tried to show that thing.
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Some one xeplain me the steps
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Give me a few minutes.
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You can write two formulas from the given data :
S1 = ut +0.5at2
S1+S2 = u(2t) +0.5a(2t)2
S1+S2 = 2ut + 2at2
Now, notice all the answers have S1, S2 and time.... Initial velocity has been eliminated. So we know we have to play around with the second equation.
Using the first equation :
ut = S1 -0.5at2
Substitute into the second one :
S1+S2 = 2*(S1 -0.5at2) + 2at2
Simplifying :
S1+S2 = 2S1 - at2 +2at2
S2-S1 = at2
(S2-S1)/t2 = a
Answer = A
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Some one xeplain me the steps
s1 = ut + 0.5at^2
u = (2s1 - at^2)/2t
2. s1+s2 = 2ut + 2at^2 (Because it takes time 2t to cover a distance of s1+s2)
u = (s1+s2-2at^2)/2t
Equate both the Us
(2s1-at^2)/2t = (s1+s2-2at^2)/2t
2t gets cancelled
2s1-at^2-s1-s2+2at^2 = 0
a = (s2-s1)/t^2
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Ah, sry, my bad.
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Thanks a million
I have soo many other doubts can yu help
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How is it C?
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Thanks a million
I have soo many other doubts can yu help
Lets say the 2m ball is Ball A and the m ball is Ball B.
Hence,
UA - -Ub = 5u/3 --u/3
I'm using the idea that when there is an elastic collision the initial velocity of body A + the initial velociity of body B is equal to the final velcoity of body B + that of Body A.
Make sure to use VELCOITIES.
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How is it C?
Sketch the wave on an axis as it would appear a few seconds later in time. Compare to the original... VOILA !!!
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Thanks a million
I have soo many other doubts can yu help
The total KE before collision, i.e 1.5mu^2, should be equal to total KE after collision.
And the momentum before collision should be the momentum after collision.
A satisfies both the conditions. Answer : A
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http://www.xtremepapers.me/forums/viewtopic.php?f=26&t=9216&start=10
How about this
Conditions for an object to be in equilbrium ?
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Well, you gotta measure the lengths, and check if R+U = W
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Kinda figure that out so the answer is A ..
N that momentum question can yu pls show the working
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Sketch the wave on an axis as it would appear a few seconds later in time. Compare to the original... VOILA !!!
Could you please elaborate a li'l more? I didn't get it..
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Could you please elaborate a li'l more? I didn't get it..
Give me some time.
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Give me some time.
Sure, thanks a lot
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I've done the sketch below. The dotted curve represents the wave a few seconds later in time.
Notice all the points except C have moved downwards.
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Kinda figure that out so the answer is A ..
N that momentum question can yu pls show the working
Total Ke b4 collision => 0.5*2m*u^2 + 0.5*m*(-u)^2 = mu^2 + mu^2/2 = 3/2 mu^2
Total momentum b4 => 2mu - mu = mu
A. KE after collision -> 0.5*2m*(u/3)^2 + 0.5*m*(5u/3)^2 = mu^2 /9 + 25mu^2/18 = 3/2 mu^2
Momentum = 2m*(-u/3) + m*(5u/3) = 5mu/3 - 2mu/3 = 3mu/3 = mu
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I've done the sketch below. The dotted curve represents the wave a few seconds later in time.
Notice all the points except C have moved downwards.
Okay, thanks, but how does it show that at that point the wave is moving UPWARDS with MAXIMUM speed?
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Okay, thanks, but how does it show that at that point the wave is moving UPWARDS with MAXIMUM speed?
If you draw a straight vertical line through point A, the point where the line intersects with the dotted curve is the point at which A is later on in time.
Notice, point A will have moved downwards.
Do the same with the rest. Only C moves upwards.
Additionally, C moves the greatest distance in the same time as the rest. Velocity = distance/time
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So yu mean to say for the above explanation we read the graph from opp direction lke from D
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http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov/
Question 13
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http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov/
Question 13
Which variant ?
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Im not sure about variant 1. The torque is obviously 4.5Nm
{Tension = Torque/diameter = 3/0.1 = 30
Torque = tension*diameter = 30*0.15=4.5}
Although, i'm not sure why the tension is 60. Maybe, because the lower belt has no tension, and for a torque two forces acting in opposite directions
are required, I guess, the tension in the lower belt is transferred to the upper belt which adds up to 60 (30+30).
The projectile one is easy
Let's say the initial velocity of the projectile is u
The velocity at highest point is horizontal with velocity ucos 45
KE = 0.5m*(ucos45)^2 = 0.5m*u^2*0.5
0.5E
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Which variant ?
Variant 11
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Im not sure about variant 1. The torque is obviously 4.5Nm
{Tension = Torque/diameter = 3/0.1 = 30
Torque = tension*diameter = 30*0.15=4.5}
Although, i'm not sure why the tension is 60. Maybe, because the lower belt has no tension, and for a torque two forces acting in opposite directions
are required, I guess, the tension in the lower belt is transferred to the upper belt which adds up to 60 (30+30).
The projectile one is easy
Let's say the initial velocity of the projectile is u
The velocity at highest point is horizontal with velocity ucos 45
KE = 0.5m*(ucos45)^2 = 0.5m*u^2*0.5
0.5E
0.5m*(ucos45)^2 = 0.5m*u^2*0.5 i gt the the answer till here how do yu simplify so that yu get 0.50E
ya i know even i gt tension as 30 But the answer is 60
idk y so askin
THANKS
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0.5m*(ucos45)^2 = 0.5m*u^2*0.5 i gt the the answer till here how do yu simplify so that yu get 0.50E
ya i know even i gt tension as 30 But the answer is 60
idk y so askin
THANKS
They've said that the kinetic energy initially is E, which is E = 0.5mu^2 (substitute this E in 0.5m*u^2*0.5)
SO u get 0.50E
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Question 12 and Question 22? http://www.xtremepapers.me/CIE/Internat ... 2_qp_1.pdf
Please explain.
THANKS
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Question 12 and Question 22? http://www.xtremepapers.me/CIE/Internat ... 2_qp_1.pdf
Please explain.
THANKS
The link is broken.
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A steel string on an electric guitar has the following properties.
diameter = 5.0 × 10^–4m
Young modulus = 2.0 × 10^11Pa
tension = 20 N
The string snaps, and contracts elastically.
By what percentage does a length l of a piece of the string contract?
A 5.1 × 10^–4%
B 5.1 × 10^–2%
C 1.3 × 10^–4%
D 1.3 × 10^–2 %
I know the Answers but i wanna Know hw do yu get them
Pls Show the workin
THANKS
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The link is broken.
thanks ari i gttta the answer
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A steel string on an electric guitar has the following properties.
diameter = 5.0 × 10^–4m
Young modulus = 2.0 × 10^11Pa
tension = 20 N
The string snaps, and contracts elastically.
By what percentage does a length l of a piece of the string contract?
A 5.1 × 10^–4%
B 5.1 × 10^–2%
C 1.3 × 10^–4%
D 1.3 × 10^–2 %
I know the Answers but i wanna Know hw do yu get them
Pls Show the workin
THANKS
is it B?
E = stress/ strain = FL/A*e
2.0 * 10^11 = 20l/(1.96*10^-7)e
e = 20l/12500pi
%change = e/l *100 = (20l/12500pi)/l *100 = 5.1*10^-2
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Yeah but how
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Yeah but how
check my previous post again (EDIT)
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is it B?
E = stress/ strain = FL/A*e
2.0 * 10^11 = 20l/(1.96*10^-7)e
e = 20l/12500pi
%change = e/l *100 = (20l/12500pi)/l *100 = 5.1*10^-2
highbury.. how did yu get e = 20l/12500pi
where did yu get 12500pi from ???
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highbury.. how did yu get e = 20l/12500pi
where did yu get 12500pi from ???
Check attachment:
Hope you get it...