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Physics: Post your doubts here!

Abu mota

Abu mota

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h4rriet said:
11. Pressure increases with depth (p=density x g x height). So does Force, because F=pA and A is constant.
18. Pressure=density x g x height. The height is the difference in height between the two tubes. The difference in height is 2h, because the liquid in the right hand tube moves h metres up and the liquid in the left hand tube moves h metres down.

14. At the highest point, its vertical velocity = o ms-1, so its K.E. can only come from its horizontal component, which is constant throughout, and is equal to xcos45.
19. p=F/A and density=mass/volume. F/g=mass and A x height=volume.
23. This is something you've just got to memorise.
25. D is correct because points equidistant from a node = to a node, and all nodes have the same displacement and all points on a wave have the same frequency. A is wrong because not all points on a stationary wave are in phase; only points in a certain segment. B is wrong because not all points have the same displacements, e.g. nodes have 0 and antinodes have maximum. C is wrong because only points in a segment are in phase.
27. An electric field is defined as the force per unit positive charge placed inside the field. That's F/q.
Click to expand...
Ty, I still don't understand 14, 19
 
S

sara kamal

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zackle09 said:
can someone solve these questions pls
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

14, 15, 31 38
and for question 32 how is it C..i thought it was D.
Click to expand...

Q14
Pivot is taken as the end of trailor and then since resultant moment =0,y is taken as the force we need to find
anticlockwise=clockwise
y x 20m=30000N x 20
so y(force)=15000N

Q15
not able to understand Q itself

Q32
I=Q/t=8/20,so I=0.4Ampere
now V=IR=0.4 x 30=12Volts

Q38
find total resistance of voltmeter and resistor connected in parallel by 2oo x 200/400=100ohms
then find the share of this resistance in total batterys voltage by :
100/500(total resistance) x 60=12V,since both voltmeter and resistor is in parallel each is having 12 V.
 
Z

zackle09

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sara kamal said:
Q14
Pivot is taken as the end of trailor and then since resultant moment =0,y is taken as the force we need to find
anticlockwise=clockwise
y x 20m=30000N x 20
so y(force)=15000N

Q15
not able to understand Q itself

Q32
I=Q/t=8/20,so I=0.4Ampere
now V=IR=0.4 x 30=12Volts

Q38
find total resistance of voltmeter and resistor connected in parallel by 2oo x 200/400=100ohms
then find the share of this resistance in total batterys voltage by :
100/500(total resistance) x 60=12V,since both voltmeter and resistor is in parallel each is having 12 V.
Click to expand...

thankyouuu :D
 
Y

Yousif Mukkhtar

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sara kamal said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf plz explain Q16.


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf plz explain Q13


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf plz explain Q8,25,27
Click to expand...
This is for s10
To find efficiency. Efficiency= Useful output/Total input
The input is the work done. INPUT ENERGY= Fs
ENERGY OUTPUT is the potential energy gained mgh

So find the height using trigonometry
height= s sin alpha.
So the efficiency:
Efficiency= mg x s sin alpha/ fs
Cancel the s you get:
mgx sin alpha/ F which is ANSWER D

This is for w10
Q8
Take XY first:
s=40, t=12, a=? and u=?

s=ut+0.5at^2
40=12u+72a
u=(40-72a)/12

v=u+at
v=(40/72a)/12 +12a

Now for YZ v from previous part will be u for the next part
s-40, u=(40/72a)/12 +12a t=6, a+?
s-ut+.5at^2
40= ((40/72a)/12 )6)+18a
In the end you will get a=0.37 ms-2



Q25
Tube X
L= lambda/4 ( Let L be length of both tubes)
Frequency is F
Tube Y
L=lambda/2
Frequnecy=?

Now as for Tube Y L is double of L of Tube X so Frequency is also doubled.
 
Y

Yousif Mukkhtar

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sara kamal said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf plz explain Q16.


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf plz explain Q13


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf plz explain Q8,25,27
Click to expand...

For q27
Step 1 .
a= (lamda xD)/x
a=(600 x 10^-9xD)/1x10^-3
a=6x10^-9 D
--------------------
Step 2
6x10^-9= (600x10^-9(D+2))/ 3x10^-3
6x10^-4=2x10^-4(D+2)
D=3-2
D-1.0m
a=6x10^-4x1.0x10^3
a=0.6mm
 
Abu mota

Abu mota

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Rahma Abdelrahman said:
And these also please...
Anyone??!!!
Click to expand...
Q. 29 the answer is c, because what causes the difference between pd and emf is always the internal resistance of the cell. B is wrong because it doesn't really make any sense, connecting wires have absolutely no effect on the emf of a cell.
Q.1 you have to arrange the momentum formula and ke formula to get ke=p2/2m,then substitute ke1 and ke2 and put them over each other, cancel p(momentum) then just solve the equation, the answer is A.
Q.2 I don't really understand the correct answer, but you're wrong because both points on the diagram are going in, meaning they are both negative, therefore they repel not attract.
 
Rahma Abdelrahman

Rahma Abdelrahman

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Abu mota said:
Q. 29 the answer is c, because what causes the difference between pd and emf is always the internal resistance of the cell. B is wrong because it doesn't really make any sense, connecting wires have absolutely no effect on the emf of a cell.
Q.1 you have to arrange the momentum formula and ke formula to get ke=p2/2m,then substitute ke1 and ke2 and put them over each other, cancel p(momentum) then just solve the equation, the answer is A.
Q.2 I don't really understand the correct answer, but you're wrong because both points on the diagram are going in, meaning they are both negative, therefore they repel not attract.
Click to expand...

Q1. I know we have to arrange it but i dont get how to do that.. how did u get ke=p2/2m?
Thanks :D
 
Tkp

Tkp

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bogus said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf
question 36 ... syed1995 h4rriet . Thanks in advance guys! :)
Click to expand...
well c the voltmeter is connected with the fixed resistor
apply the formlua of potentiometer
p/p+q*emf(where p is the fixed resistor resistance and q is the variable resistor)
so increasing the q decreases the voltage so voltage decrases
and the ammmeter is directly connected with the fixed resistor so there is no change in readings.Hope it helped u:)
 
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