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Physics: Post your doubts here!

G

geek101

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bamteck said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf

HELP NEEDED for no. 6 (b) :(
Its too confusing ! pff
Click to expand...

the formula for Power = R^2 / V were gonna use this for each scenario
so just keep in mind that each resistor is 38.4 ohms and the V is gonna be 240 V no matter what.

1) S1 the control switch is off >> R = 0 ohms >> P = 0 W

2) R in B is zero because it parallel to the wire (when S2 is closed) and the wire has negligible resistance and so when resistors are in parallel the total R is lesser than the least, so R in B is o. and R in C is obviously 0.
so R total = 38.4 ohms (of A)
P = 240^2 / 38.4 = 1.5 kW

3) the R of A and B will be taken here...now its a series circuit with A, B and the power supply. eliminate C.
total R = 38.4 + 38.4
P = 240^2 / 38.4 x 2 = 3 kW

4) All the components resistance are to be taken...
r total :
38.4 + 38.4 = 76.8 ohms >> A + B, now these are in parallel with C so
78.6 x 38.4 / 76.8 + 38.4 = 25.6 ohms
P = 240^2 / 25.6 = 2.25 kW

hope that helps =)
 
bamteck

bamteck

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geek101 said:
the formula for Power = R^2 / V were gonna use this for each scenario
so just keep in mind that each resistor is 38.4 ohms and the V is gonna be 240 V no matter what.

1) S1 the control switch is off >> R = 0 ohms >> P = 0 W

2) R in B is zero because it parallel to the wire (when S2 is closed) and the wire has negligible resistance and so when resistors are in parallel the total R is lesser than the least, so R in B is o. and R in C is obviously 0.
so R total = 38.4 ohms (of A)
P = 240^2 / 38.4 = 1.5 kW

3) the R of A and B will be taken here...now its a series circuit with A, B and the power supply. eliminate C.
total R = 38.4 + 38.4
P = 240^2 / 38.4 x 2 = 3 kW

4) All the components resistance are to be taken...
r total :
38.4 + 38.4 = 76.8 ohms >> A + B, now these are in parallel with C so
78.6 x 38.4 / 76.8 + 38.4 = 25.6 ohms
P = 240^2 / 25.6 = 2.25 kW

hope that helps =)
Click to expand...

Parallel ? How ?
Couldn't a diagram be attached please ? :(
Thanks ...
 
emkay

emkay

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bamteck said:
Parallel ? How ?
Couldn't a diagram be attached please ? :(
Thanks ...
Click to expand...

uhm! let me try?

At first, you calculate the resistance of one heating element by R=V^2/P which is 38.4 Ohms.

when switch S1 is opened the current will not be able to complete its cycle and hence Power= Zero.

when S1 and S2 are closed but S3 is opened we will neglect the resistor (B) as current is unable to pass through it. By keeping in mind that Power is always divided in the series circuit if the resistors are identical. Power in (A) and (B) would be 3.okW/2 giving us 1.5kW.

when S1, S2 and S3 are all closed, Power in (A) and (B) as calculated before was 1.5kW + Power in (C) which is also 1.5kW, all together making it 3.0kW (you need to recall that Power is added in parallel series of identical resistors)

when S1 is closed with S2 and S3 opened, as you know in series the resistors are added (i.e Rt=R1+R2+R3..) we would add the Resistor (A) and (B) together, 38.4+38.4=76.8. now use Power=V^2/R (R=76.8) and you will get 0.75kW.

when S1 and S3 are closed with S2 opened, you will be using all the resistors (A),(B) and (C). Neglect S2 it wouldn't make any difference here.
(Rt*Rc)/(Rt+Rc) that is (76.8*38.4)/(76.8+38.4) which gives 25.6 now use P=V^2/R by taking 240^/25.6 and you will get 2.25kW.
 
emkay

emkay

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Silent Hunter said:
What shows that the spacing in liquids and solids is the same?
Click to expand...
uhm, the spacing? well solids are clumped together and constantly vibrate next to one another. They tend to maintain a shape and a volume, where as liquids constantly flow over each other when they are poured. They tend to maintain a shape that is similar to their container (e.g glass or bowl) and they also have a definite volume.
 
bamteck

bamteck

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emkay said:
uhm! let me try?

At first, you calculate the resistance of one heating element by R=V^2/P which is 38.4 Ohms.

when switch S1 is opened the current will not be able to complete its cycle and hence Power= Zero.

when S1 and S2 are closed but S3 is opened we will neglect the resistor (B) as current is unable to pass through it. By keeping in mind that Power is always divided in the series circuit if the resistors are identical. Power in (A) and (B) would be 3.okW/2 giving us 1.5kW.

when S1, S2 and S3 are all closed, Power in (A) and (B) as calculated before was 1.5kW + Power in (C) which is also 1.5kW, all together making it 3.0kW (you need to recall that Power is added in parallel series of identical resistors)

when S1 is closed with S2 and S3 opened, as you know in series the resistors are added (i.e Rt=R1+R2+R3..) we would add the Resistor (A) and (B) together, 38.4+38.4=76.8. now use Power=V^2/R (R=76.8) and you will get 0.75kW.

when S1 and S3 are closed with S2 opened, you will be using all the resistors (A),(B) and (C). Neglect S2 it wouldn't make any difference here.
(Rt*Rc)/(Rt+Rc) that is (76.8*38.4)/(76.8+38.4) which gives 25.6 now use P=V^2/R by taking 240^/25.6 and you will get 2.25kW.
Click to expand...

Thanks a lot :)
 
emkay

emkay

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bamteck said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

URGENT HELP NEEDED for no. 2 (d) & 7 (b) :(
Click to expand...

Since it's urgent, let's have a look.
For Question# 2(d)

the question starts from "deduce" means we have to show some working before writing the comments.
Find the Ek (Kinetic Energy) for the ball B before and after the collision.

Before Collision
Ek (for B) = 1/2*mv^2 which is 1/2*1.2*(4)^2 and gives 9.6 J. (i took the velocity 4ms-1 from the graph)

After Collision
Ek (for B) = 1/2*mv^2 which is 1/2*1.2*(2.89)^2 and gives 4.89 J. (again, i took the velocity from the graph)

In any collision, the linear momentum is always conserved where as the Ek may not. if the Ek after the collision is less than the Ek before the collision it is said to be inelastic but if the Ek after the collision is greater than the Ek before the collision than it is due to chemical electric energy (e.g an explosive between the two sphere's), if both of the Ek's are same then it is said to be an elastic collision.
Now, as you can see that the Ek after the collision is less than the Ek before the collision, therefore it is inelastic.

For Question#7(b)

Draw the arrow heads following their route for every current you see so it makes a much clear picture in you mind.
For (i) The current I1 and I2 meet at the point B, where as the current I3 draws away to point X.
so it should be I3 = I1+I2.
For (ii) Now focus the loop BCXYB, take a sharp look starting for the E2 then going to the point C,B,Y,X and back to E2.
it should be, E2 = I3R + I2R
For(iii) it is the simplest one, the two cells are connected in the same direction so it will be E1-E2 and the rest is the path of currents (I1,I2,I3) which multiplies with respective resistors. here is how it goes, I1R - I2R + I1 R which means, 2I1R-I2R.
so it is, E1-E2 = 2I1R-I2R.
 
emkay

emkay

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axetreme.O said:
My simple question is-" Can light apply force to something?" Kindly post your opinions..however facts are more welcome....:whistle:
Click to expand...

Uhm, this actually gave me a strike. :p

okay, for light to have force it must have mass and acceleration. because force=m*acc.
but that is not the case here, why? let's see.
the best way to describe force of light is the absorption and emission of light. when light is absorbed so is it's energy and momentum the same goes for the emission of light. for light to have a force, mass is not necessary but it must have momentum. (i don't get more than that :p) i think to go further quantum mechanics is what's required.
 
G

geek101

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bamteck said:
Parallel ? How ?
Couldn't a diagram be attached please ? :(
Thanks ...
Click to expand...

first of all i skipped scenario 4 there...sorry here are the answers with the diagrams..
I removed every branch with an open switch...and if the switch was closed i made it part of the wire...
and the first one doesnt really need a diagram. hope its not too tiny :p
 

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bamteck

bamteck

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emkay said:
Since it's urgent, let's have a look.
For Question# 2(d)

the question starts from "deduce" means we have to show some working before writing the comments.
Find the Ek (Kinetic Energy) for the ball B before and after the collision.

Before Collision
Ek (for B) = 1/2*mv^2 which is 1/2*1.2*(4)^2 and gives 9.6 J. (i took the velocity 4ms-1 from the graph)

After Collision
Ek (for B) = 1/2*mv^2 which is 1/2*1.2*(2.89)^2 and gives 4.89 J. (again, i took the velocity from the graph)

In any collision, the linear momentum is always conserved where as the Ek may not. if the Ek after the collision is less than the Ek before the collision it is said to be inelastic but if the Ek after the collision is greater than the Ek before the collision than it is due to chemical electric energy (e.g an explosive between the two sphere's), if both of the Ek's are same then it is said to be an elastic collision.
Now, as you can see that the Ek after the collision is less than the Ek before the collision, therefore it is inelastic.

For Question#7(b)

Draw the arrow heads following their route for every current you see so it makes a much clear picture in you mind.
For (i) The current I1 and I2 meet at the point B, where as the current I3 draws away to point X.
so it should be I3 = I1+I2.
For (ii) Now focus the loop BCXYB, take a sharp look starting for the E2 then going to the point C,B,Y,X and back to E2.
it should be, E2 = I3R + I2R
For(iii) it is the simplest one, the two cells are connected in the same direction so it will be E1-E2 and the rest is the path of currents (I1,I2,I3) which multiplies with respective resistors. here is how it goes, I1R - I2R + I1 R which means, 2I1R-I2R.
so it is, E1-E2 = 2I1R-I2R.
Click to expand...

Thanks a lot man :)
But why n the question, its says 'use your answer in (c)' ? oO'
 
hussamh10

hussamh10

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bamteck said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf

HELP NEEDED for no. 6 (b) :(
Its too confusing ! pff
Click to expand...
First of all find the resistance of the individual resistor which is 38.4
First is open close close so
No current as the first switch is open so no power
then Close Close open
Current passes but as Switch two is open so Current bypasse the resistor B so the resistance is not taken into account of B see because the current has an easier route so we say it is a short circuit so when we take the effective resistance it wiill only be 38.4 as Switch 3 is openso
p=V^2/Rand we get 1500 watts
In Third Case it is Close CLOSE CLOSE
Now againn the current bypasse resistance B a short circuit occur so we take the resistance of A and C in parrallel as in the circuit.
Hope you get the idea if not please feel free to ask please:)
 
hussamh10

hussamh10

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Silent Hunter said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf

q(3) (b)(iii) and (c)

q(5)(d)(i)(ii)



thanks alot
Click to expand...
Vertical height is 8.6 by
2as=v^2-u^2 V=15cos60 u=0
so we get s=8.6(vertical height)
So we find out the time As verticla component of velocity is zero at top Highest point so
s=ut+.5x9.81xt^2
8.6=(0)t=.5x9.81xt^2
we get T=1.32
So horizontal component is 6.15/1.32=4.64m/s
In c part the important thing is that before hitting the wall the ball has horizontal component while vertical component is zero so now u can find the change
Q5 resistance is zero so pwer is zero obvious and in infinite resistance no current passes so zero rsistance i guess...
 
C

caro24

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caro24 said:
plz can someone help me..what are the experiments that i can be asked in exams for nuclear physics p2,,and how can i describe them plz plz ......i really need your help..god bless u
Click to expand...
is there nobody to help me..im begging
 
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