Physics: Post your doubts here!

[gary221]

A metal cube of side l is placed in a vice and compressed elastically by two opposing forces F.
F F l
l
metal cube
How will ∆l, the amount of compression, relate to l?
A ∆l ∝ 2
1
l
B ∆l ∝
l
1
C ∆l ∝ l D ∆l ∝ l
2

paper link??

 

[selena]

paper link??

winter 2011

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_2.pdf
Someone please help me with question 7 b(i), I don't understand the marking scheme.
Thanks

 

[gary221]

Aoa wr wb

Can someone please explain 4b ii of this paper?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_41.pdf

Since both the nuclei have initial kinetic energy = Ek ---> total initial kinetic energy = 2Ek.
Now, as the nuclei(both positively charged) move closer together, they gain electric potential energy.
But, as the distance btw them is being reduced, they lose gravitational potential energy.
ie kinetic energy + gravitational potential energy is lost.
electric potential energy is gained.
So, 2Ek + ΔEgpe = ΔEepe ------> 2EK= ΔEepe – ΔEgpe
So, 2Ek = (6.06 × 10–14 ) - (1.93 × 10–49 )
Ek = 3.03 * 10^-14 J
Converting into MeV,
Ek = (3.03 * 10^-14)/(1.6 * 10^-13)
Ek = 0.19 MeV.
Hope u gt it!

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_2.pdf
Someone please help me with question 7 b(i), I don't understand the marking scheme.
Thanks

Ah .. they say that 1.1*10^5 J of energy was dissipated in the 45 ohm resistor .. how much energy will be dissipated in the 15 ohm resistor..

P= I^2 * R

P directly propotional to R

45 ohms produced 110 kJ 15 ohms will produce what?

45 -- 1.1 * 10^5
15 -- x

45x = 15 * 1.1 * 10^5
x = 3.67 * 10^4 Answer

 

[A star]

Ah .. they say that 1.1*10^5 J of energy was dissipated in the 45 ohm resistor .. how much energy will be dissipated in the 15 ohm resistor..

P= I^2 * R

P directly propotional to R

45 ohms produced 110 kJ 15 ohms will produce what?

45 -- 1.1 * 10^5
15 -- x

45x = 15 * 1.1 * 10^5
x = 3.67 * 10^4 Answer

dude solve the question i posted in chem thread pls

 

[Soldier313]

thanks alot for your reply
iam a bit still confused like isnt intersity the no. of photons travelling? so if thet are constant wouldn't the current remain constant?

you're welcome anytime.
as far as i know intensity is the total energy carried by the photons. Hence, inorder to maintain a constant intensity, we must reduce the number of photons, and hence the number of photoelectrons will be reduced)
Hope yu get it inshaAllah.

 

[Soldier313]

Since both the nuclei have initial kinetic energy = Ek ---> total initial kinetic energy = 2Ek.
Now, as the nuclei(both positively charged) move closer together, they gain electric potential energy.
But, as the distance btw them is being reduced, they lose gravitational potential energy.
ie kinetic energy + gravitational potential energy is lost.
electric potential energy is gained.
So, 2Ek + ΔEgpe = ΔEepe ------> 2EK= ΔEepe – ΔEgpe
So, 2Ek = (6.06 × 10–14 ) - (1.93 × 10–49 )
Ek = 3.03 * 10^-14 J
Converting into MeV,
Ek = (3.03 * 10^-14)/(1.6 * 10^-13)
Ek = 0.19 MeV.
Hope u gt it!

Yes i understood! Thank you so much bro! JazakAllah khair .

 

[abruzzi]

Ah .. they say that 1.1*10^5 J of energy was dissipated in the 45 ohm resistor .. how much energy will be dissipated in the 15 ohm resistor..

P= I^2 * R

P directly propotional to R

45 ohms produced 110 kJ 15 ohms will produce what?

45 -- 1.1 * 10^5
15 -- x

45x = 15 * 1.1 * 10^5
x = 3.67 * 10^4 Answer

Damn that was simple.
Thank you

 

[megamind]

thank you KurayamiKimmi but can space waves go aroun the world with no satellite
https://www.xtremepapers.com/community/members/kurayamikimmi.5698/

 

[megamind]

i dont know that too
anyone who can reply this

re-posting again

 

[gary221]

re-posting again

tht was answered... the young's experiment na??

 

[Monojit Saha]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
why Q5b(ii) which region is decreasing and which is constant and why. what if region is increasing than how would the diagram be

the magnitude or strength of fields depends upon the separation between the lines.. if the separation is increasing then the strength of the field is decreasing and if the separation between the fields are decreasing, the field is increasing. When the field is approaching the earthed plate, u can c that the field lines are uniform and thus the magnitude of the field is constant as lines are parallel.. but when fields are coming out of the positive end, they are increasing separation and thus are decreasing in strength or magnitude..

 

[hellangel1]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
Question 2B

 

[1357913579]

you're welcome anytime.
as far as i know intensity is the total energy carried by the photons. Hence, inorder to maintain a constant intensity, we must reduce the number of photons, and hence the number of photoelectrons will be reduced)
Hope yu get it inshaAllah.

thanks alot
I totally got understood it
but the problem that is arising is that if you say that intensity is the total energy carried by the photon then it means if you increase the intensity you can make photo electron emission possible without reaching the thresohold frequency

 

[KurayamiKimmi]

thank you KurayamiKimmi but can space waves go aroun the world with no satellite

they can work without the satellite but they can only propagate by line of sight , which means they only travel in a straight line from transmitter to receiver .

 

[daredevil]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_22.pdf

hey guys plz c Q7b
What is MeV?? In which part of the sylabus is it mentioned if it even is in our syllabus... We hav done nothing of the sort in our class.. plz help :O

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf

hey guys plz c Q7b
What is MeV?? In which part of the sylabus is it mentioned if it even is in our syllabus... We hav done nothing of the sort in our class.. plz help :O

MeV is just another unit for energy
1 MeV = 1.6 * 10^-13 J

 

[snowbrood]

MeV is just another unit for energy
1 MeV = 1.6 * 10^-13 J

where did u get 1.6*10^-13 its not given in the question or anywhere else in the question paper so ?

 

[gary221]

where did u get 1.6*10^-13 its not given in the question or anywhere else in the question paper so ?

my teacher taught me tht...
it isnt thr in d book??