Physics: Post your doubts here!

[daredevil]

Could sb please tell me why, when generating electrical power, the electrical power output would be less than the power available from a source such as the wind or the sun. How would power be lost

power is wasted in the form of heat energy, sound energy, light energy, etc.
it is sometimes case-by-case basis so do see whether the method or process you are talking about does involve these energy changes or not.

Actually nothing in the practical world is 100% efficient. Some energy is always lost in mostly at least one of the mentioned forms, so when speaking practically power output is always less than power input.

 

[asma tareen]

Can anyone explain kirchoff's laws :/

 

[reachonlyme]

Any doubts in O levels(math,physics and chemistry),u can get help..
Look at the free O levels Statistics class
just follow the link
http://www.wiziq.com/course/40536-free-igcse-math

 

[shazmina]

Can someone help me with Qs no 17 of this paper
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf
& Qs no 14 of this paper
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf
Thanks in Advance

 

[daredevil]

Can someone help me with Qs no 17 of this paper
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf
& Qs no 14 of this paper
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
Thanks in Advance

Q.17
E(x)/E = [(2m*v^2)/2] / [(m(2v)^2)/2)]
= [mv^2]/[2mv^2]
= 1/2

So X has half the kinetic energy of Y.
answer : A

 

[shazmina]

Q.17
E(x)/E = [(2m*v^2)/2] / [(m(2v)^2)/2)]
= [mv^2]/[2mv^2]
= 1/2

So X has half the kinetic energy of Y.
answer : A

Thanks alot sis

 

[daredevil]

Thanks alot sis

no problem

 

[periyasamy]

Hi guys .Anyone can help me with this question.Thank u a lot.

 

[shazmina]

Can someone help me with Qs no 34 of this paper
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
Thanks in advance

 

[Snowysangel]

Hi guys .Anyone can help me with this question.Thank u a lot.

Isn't it mgh= (6x10^24)*(1.3x10^4)*9.81 ?

 

[Kim Shaw]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q.10
the answer is D. but why?? shouldn't it be u1-u2 since they are in opposite direction??

The collision is perfectly elastic. hence, the relative speed of approach=relative speed of separation.
Imagine 2 bodies of equal masses (B1 and B2) approaching each other...and on collision move away in opposite directions.
if B1 moves 10 steps towards B2, and B2 moves 5 steps towards B1, then they approached by 5+10=15steps
thus rel.speed of approach is u1+u2

similarly, if after collision, B1 moves towards B2 by 1 step, and B2 moves away from B1 by 1000 steps, then both bodies are separated from each other by 1000-1=999 steps.
thus rel.speed of separation is V2-V1
put both together and tada!
u1+u2=v2-v1

 

[Kim Shaw]

Can someone help me with Qs no 34 of this paper
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Thanks in advance

R=(pl)/A
A=L^2
When A is substituted in 1st equation, R=p/l
now V=l^3
l=(l^3)^(1/3)
thus, l=V^(1/3)
replacing this in 2nd equation... R=p/(V^(1/3))
answer C

 

[periyasamy]

Isn't it mgh= (6x10^24)*(1.3x10^4)*9.81 ?

Thanks buddy,but the answer is 4.19*10^7.Your solution gives me answer of 7.6*10^29.

 

[shazmina]

R=(pl)/A
A=L^2
When A is substituted in 1st equation, R=p/l
now V=l^3
l=(l^3)^(1/3)
thus, l=V^(1/3)
replacing this in 2nd equation... R=p/(V^(1/3))
answer C

Thanks you sis

 

[***amd***]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Q1 ci)
Q2 b)
Q5 di) dii)
@cardude456, @Afzal1, @Kumkum, @shazmina anyone?

 

[Kumkum]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Q1 ci)
Q2 b)
Q5 di) dii)
@cardude456, @Afzal1, @Kumkum, @shazmina anyone?

1 ci) since your're asked to find the change in gravitational potential, use this formula: -GM/r
so it'll be:
(6.67*10^-11 * 6*10^24) / (6.4*6 - 1.3*10^7) = -6.06*10^7 (i'm not sure about the answer)
the radius's have been changed to meters

2 b) not so sure, sorry

5 di) since it's an alpha particle it will deflect upwards, but not so much.
dii) since it's a beta particle, it will deflect downwards. the deflection will be large.
i think it's like this:


hope u got it

 

[Kim Shaw]

Thanks you sis

No prob

 

[Kim Shaw]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Q1 ci)
Q2 b)
Q5 di) dii)
@cardude456, @Afzal1, @Kumkum, @shazmina anyone?

for question 2b...
ok, so final temperature is mpt of gold, which is 1340
also, 5% of mixture will be silver, so 95% will be gold
formula used heat=mcdt, during change in state=mlf
dT(change in temp)= (whatever very high temp of gold which we don't knw of) - 1340
the heat lost by the gold=the heat gained by the silver(given that no heat is lost to the surroundings)
heat lost by gold=(0.95m x 129 x dT)
heat gained by silver: so we have to heat silver till it is solid at its mpt. then heat more until it is liquid at its mpt. THEN heat more for it to reach mpt of gold
therefore, heat gained by silver=(o.o5m x 235 x (1240-(27+273.15))) + (0.05m x 105 x 1000) + (0.05m x 235 x (1340-1240))
equate both these and m will cancel out, and tada!! dT=143K(3sf)
dT=initial temp of gold when silver was added-1340
thus ans=143+1340=1483
and here, dear friend, is the most simple and comprehensive way that i could explain a question like that. But don't hesitate to ask anything u don't understand in what i've scribbled up top

 

[periyasamy]

Hi guys.Can anyone solve my problem here.For the picture i post below,did i draw it rightly in terms that there is only one force acting between moon n earth n that is ONLY an attractive force

 

[periyasamy]

Hi guys.Can anyone solve my problem here.For the picture i post below,did i draw it rightly in terms that there is only one force acting between moon n earth n that is ONLY an attractive force

Sorry guys i just google searched n saw this picture but i dont reallly understand why it is saif that the force acting for moon is opposite for the force acting on earth.Anybody can help me.Thanks.