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Physics: Post your doubts here!

Cute Angella

Cute Angella

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can someone pls help me out with this...
M/J/06 P4 q2 b(ii)

and O/N/07 P4 q3 b(ii) 2
how can we use the graph in this question...i got the answer by using the formula KE=1/2mw2 A2...but i dont know how to use the graph..???
 
Cute Angella

Cute Angella

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O/N 06 P4 q7 b (phy A2)
what i know is that frequency of light hv no effect on the rate of emission of electrons.... rate only depends on the intensity of light.... n here light intensity is constant.... so how does the rate decreases..???? n how the photons r fewer per second at higher frequency ???????
 
Cute Angella

Cute Angella

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mahabaloch said:
anyone can help me with nov 07 q5 part c plz
Click to expand...
first find the initial energy stored in the capacitor using the initial voltage...
E=1/2 CV^2
E=1/2 x 220 x 10^-6 x 15^2
E=0.02475 J
since half of this energy is lost so only half is left behind
so divide this energy by 2 to get half of it
E=0.012375 J
now with this left over energy find the voltage using the same above formula
E=1/2 CV^2
0.012375=1/2 x 220 10^-6 x V^2
V=10.6 V ...
hope u get it .. :)
 
S

shaheerrn

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BreakingBad said:
Stress is force per unit area. The area is a hundred times smaller so A probably. Besides, options B,C and D are absurd. It would make more sense if those numbers were more like 1000 times, 10 times etc.


i) Already been answered.
ii)
P.d. across the thermistor and the 5000 ohms resistor combined is 6-3.6 = 2.4V
The current in the circuit is V/R = 3.6/2000 = 0.0018A

Therefore, combined resistance for the thermistor and the 5000 ohms resistor is V/I = 2.4/0.0018 = 1333.333 ohms

1/RT = 1/Rt + 1/Rr
1/1333.333 = 1/Rt + 1/5000
Rt = 1820 ohms
Sorry but I haven't been looking at most of the convos lately.[/quote

no these are powers A.10^0 B.10^1 C 10^2 D. 10^3
and the answer is B
Click to expand...
 
Cute Angella

Cute Angella

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Salman917 said:
how will we understand wen to use mv^2/r=GMm/r^2 N wen 2 use 0.5mv^2/r=GMm/r^2???????????????????
Click to expand...
i don't think we ever use the second equation..!!!!
we always use the first one where centripetal force is provided by the gravitational force.....
 
B

BreakingBad

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Salman917 said:
how will we understand wen to use mv^2/r=GMm/r^2 N wen 2 use 0.5mv^2/r=GMm/r^2???????????????????
Click to expand...
When you are relating forces, you use mv^2/r because centripetal F = mv^2/r and Gravitational F = GMm/r^2
When you are relating energies, you use 1/2mv^2 because (K)E = 1/2mv^2 and GPE = -GMm/r (You don't equate the 2 though, usually)
 
Cute Angella

Cute Angella

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can someone pls help me out with a few questions of phy A2.....
1. M/J/06 P4 q2 b(ii)
link =http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s06_qp_4.pdf

2. O/N/07 P4 q3 b(ii) 2
how can we use the graph in this question...i got the answer by using the formula KE=1/2mw2 A2...but i dont know how to use the graph..???
link =http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_4.pdf

3. O/N 06 P4 q7 b
what i know is that frequency of light hv no effect on the rate of emission of electrons.... rate only depends on the intensity of light.... n here light intensity is constant.... so how does the rate decreases..???? n how the photons r fewer per second at higher frequency ???????
link =http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_4.pdf

4. M/J/09 P4 q9 b
can someone pls explain why the source must decay by 8% ???
link =http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf

pls someone do help..!!!
 
Last edited:
Salman917

Salman917

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BreakingBad said:
When you are relating forces, you use mv^2/r because centripetal F = mv^2/r and Gravitational F = GMm/r^2
When you are relating energies, you use 1/2mv^2 because (K)E = 1/2mv^2 and GPE = -GMm/r (You don't equate the 2 though, usually)
Click to expand...
Thanx a lot! :)
 
B

BreakingBad

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Cute Angella said:
can someone pls help me out with a few questions of phy A2.....
1. M/J/06 P4 q2 b(ii)
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s06_qp_4.pdf

2. O/N/07 P4 q3 b(ii) 2
how can we use the graph in this question...i got the answer by using the formula KE=1/2mw2 A2...but i dont know how to use the graph..???
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_4.pdf

3. O/N 06 P4 q7 b
what i know is that frequency of light hv no effect on the rate of emission of electrons.... rate only depends on the intensity of light.... n here light intensity is constant.... so how does the rate decreases..???? n how the photons r fewer per second at higher frequency ???????
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w06_qp_4.pdf

4. M/J/09 P4 q9 b
can someone pls explain why the source must decay by 8% ???
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s09_qp_4.pdf

pls someone do help..!!!
Click to expand...
Physics' links please :p
 
AbbbbY

AbbbbY

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Cute Angella said:
can someone pls help me out with a few questions of phy A2.....
1. M/J/06 P4 q2 b(ii)
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s06_qp_4.pdf

2. O/N/07 P4 q3 b(ii) 2
how can we use the graph in this question...i got the answer by using the formula KE=1/2mw2 A2...but i dont know how to use the graph..???
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w07_qp_4.pdf

3. O/N 06 P4 q7 b
what i know is that frequency of light hv no effect on the rate of emission of electrons.... rate only depends on the intensity of light.... n here light intensity is constant.... so how does the rate decreases..???? n how the photons r fewer per second at higher frequency ???????
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_w06_qp_4.pdf

4. M/J/09 P4 q9 b
can someone pls explain why the source must decay by 8% ???
link = http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Biology (9700)/9700_s09_qp_4.pdf

pls someone do help..!!!
Click to expand...

You do realize that you've linked Bio papers in a Physics thread, right?
 
M

mahabaloch

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Cute Angella said:
first find the initial energy stored in the capacitor using the initial voltage...
E=1/2 CV^2
E=1/2 x 220 x 10^-6 x 15^2
E=0.02475 J
since half of this energy is lost so only half is left behind
so divide this energy by 2 to get half of it
E=0.012375 J
now with this left over energy find the voltage using the same above formula
E=1/2 CV^2
0.012375=1/2 x 220 10^-6 x V^2
V=10.6 V ...
hope u get it .. :)
Click to expand...

Thank u
 
Cute Angella

Cute Angella

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ooo m so sorry... actually i was solving bio past papers.. n didn't realize that by mistake i linked that to phy...:p
now i hv changed the links... sorry once again....
 
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