As physics p1 MCQS YEARLY ONLY.

[magnesium]

i need 2003 nov solution? plz

 

[syed1995]

I think nobody can solve this question in MCQ, go to nov 2002 q.36. What the hell is that?

We Just find the PD dropped on each of X and Y and find the difference...

The answer is coming as 0.67V.

The PD will be.. Reading at Y - Reading at X.

since Y is after 2 Resistors... the resistance will be 4/3 V .. and at X there is only one resistor.. so Resistance is 2/3.

Y-X .. = 4/3 - 2/3 = 2/3 Answer.

 

[TaffsAsLevel]

We Just find the PD dropped on each of X and Y and find the difference...

The answer is coming as 0.67V.

The PD will be.. Reading at Y - Reading at X.

since Y is after 2 Resistors... the resistance will be 4/3 V .. and at X there is only one resistor.. so Resistance is 2/3.

Y-X .. = 4/3 - 2/3 = 2/3 Answer.

I got the answer already, but thanks a ton <3

 

[minie23]

We Just find the PD dropped on each of X and Y and find the difference...

The answer is coming as 0.67V.

The PD will be.. Reading at Y - Reading at X.

since Y is after 2 Resistors... the resistance will be 4/3 V .. and at X there is only one resistor.. so Resistance is 2/3.

Y-X .. = 4/3 - 2/3 = 2/3 Answer.

Can you help with w11 p12 no. 15 ? please

 

[h4rriet]

Can you help with w11 p12 no. 15 ? please

May I? mg-kv=o because the velocity is constant. mg=kv. Rearrange the formula to find v, then use 1/2m(v^2) to find the K.E.

 

[h4rriet]

i need 2003 nov solution? plz

Question what?

 

[minie23]

May I? mg-kv=o because the velocity is constant. mg=kv. Rearrange the formula to find v, then use 1/2m(v^2) to find the K.E.

oO'
didnt get it

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
No. 15

 

[syed1995]

Can you help with w11 p12 no. 15 ? please

W=Fs
Fs=Change in Kinetic Energy..
Fs=4

When Both the F and s are multiplied by 2... we get..

Change In Kinetic Energy
=2F*2s
=4Fs
=4(4)
Change in KE = 16J..

Initial was 4J.. so 16+4 .. Answer is 20J.

 

[h4rriet]

oO'
didnt get it

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
No. 15

mg is the weight, so the force downwards. kv is the air resistance, so the force upwards. The velocity is constant, so there's no resultant force. If you add the two forces, you must get 0. mg+-kv=o. +-=-.

15. WD=Fs. If Fs=2, then 2F2s=x. Cross multiply and you'll get 16. Add that to 4 to get 20.

 

[minie23]

W=Fs
Fs=Change in Kinetic Energy..
Fs=4

When Both the F and s are multiplied by 2... we get..

Change In Kinetic Energy
=2F*2s
=4Fs
=4(4)
Change in KE = 16J..

Initial was 4J.. so 16+4 .. Answer is 20J.

mg is the weight, so the force downwards. kv is the air resistance, so the force upwards. The velocity is constant, so there's no resultant force. If you add the two forces, you must get 0. mg+-kv=o. +-=-.

15. WD=Fs. If Fs=2, then 2F2s=x. Cross multiply and you'll get 16. Add that to 4 to get 20.

Thanks to both of you ?
Why should we add 4 ?

 

[h4rriet]

Thanks to both of you ?
Why should we add 4 ?

Because initially it was 4 and then it became 6. That means an additional 2 J was added to the 4.

 

[magnesium]

Question what?

http://www.freeexampapers.com/#A Level/Physics/CIE/2002 JunQ 11, 21, 35 PLZ!

 

[ahmed abdulla]

We Just find the PD dropped on each of X and Y and find the difference...

The answer is coming as 0.67V.

The PD will be.. Reading at Y - Reading at X.

since Y is after 2 Resistors... the resistance will be 4/3 V .. and at X there is only one resistor.. so Resistance is 2/3.

Y-X .. = 4/3 - 2/3 = 2/3 Answer.

Q1 A box of mass 8.0 kg rests on a horizontal, rough surface. A string attached to the box passes
over a smooth pulley and supports a 2.0 kg mass at its other end.
When the box is released, a friction force of 6.0 N acts on it.
What is the acceleration of the box?
answer is 1.4ms-2
can u solve it via newton second law plz?

Q2--

An electric power cable consists of six copper wires c surrounding a steel core s
1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100 Ω.
What is the approximate resistance of a 1.0 km length of the power cable?
B 1.6 Ω

 

[PhyZac]

i need 2003 nov solution? plz

 

[magnesium]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdfQ16/

 

[Lyfroker]

i need 2002 nov ms

 

[magnesium]

i need 2002 nov ms

http://www.freeexampapers.com/#A Level/Biology/CIE/2002 Nov

 

[Lyfroker]

http://www.freeexampapers.com/#A Level/Biology/CIE/2002 Nov

thats bio...

 

[magnesium]

Lyfroker
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdfQ16 CAN u help

 

[magnesium]

thats bio...

http://www.freeexampapers.com/#A Level/Physics/CIE/2002 Nov