Physics: Post your doubts here!

[emkay]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf

Asalaomalikum ... need help in Q4(b)

Thank you

Ah! let's see this, you're going to use the formula Ek = 1/2 mv^2 here.
use the kinetic energy you proved in (ii) which was found by Ek = Q.V putting Q=1.66*10^-19 and V= 250, by solving such values you will prove that Ek = 4.0*10^-17
now in part (iii) all you have to do is put this Ek that you found in part (ii) and put m= rest mass of an electron as 9.11*10^-31 and you can easily find the value of v. which should be 9.4*10^6 m/s.

 

[Silent Hunter]

Ah! let's see this, you're going to use the formula Ek = 1/2 mv^2 here.
use the kinetic energy you proved in (ii) which was found by Ek = Q.V putting Q=1.66*10^-19 and V= 250, by solving such values you will prove that Ek = 4.0*10^-17
now in part (iii) all you have to do is put this Ek that you found in part (ii) and put m= rest mass of an electron as 9.11*10^-31 and you can easily find the value of v. which should be 9.4*10^6 m/s.

thank you but i was asking for 4(b)

 

[queen of the legend]

thank you but i was asking for 4(b)

since W=QV for work done in electric circuit and kinetic energy K= 0.5 mv^2 which is also equal to work done we can relate these two equations as
QV=0.5mv^2
the question says that the pd is the same and ofcource Q is also constant so v, the speed will be always same as well
i hope you understood!

 

[Silent Hunter]

since W=QV for work done in electric circuit and kinetic energy K= 0.5 mv^2 which is also equal to work done we can relate these two equations as
QV=0.5mv^2
the question says that the pd is the same and ofcource Q is also constant so v, the speed will be always same as well
i hope you understood!

Thank you very much

 

[xyz!]

COULD SOMEONE ON EARTH PLEASE POST THE END-OF-CHAPTER ANSWERS OF PHYSICS COURSE BOOK?
I AM FINDING THEM FROM YEARS AND WITH NO LUCK! IF ANY ONE HAS TEACHER'S C-D ROM, HE MIGHT HAVE ACCESS TO THAT- AS FAR AS I HAVE HEARD.

i hav dem for A level chptrs..u want AS or A?

 

[xyz!]

hiee..dis is a doubt regarding paper 5
here's the link..
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_51.pdf
can sum1 pls the diagram(a labelled one pls) fr the first question and attach it..?? i read the ms but m not able to undrstnd!

 

[Minato112]

12 Magnesium chloride (MgCl2) and calcium oxide (CaO) are ionic compounds.
Electricity is passed through the molten compounds, in two separate experiments.
Which are the products at the cathodes?
a magnesium and calcium
b magnesium and oxygen
c oxygen and calcium
d chlorine and oxygen

Please post chemistry questions in respective chemistry threads.

http://www.xtremepapers.com/community/threads/chemistry-post-your-doubts-here.9859/

 

[raamish]

COULD SOMEONE ON EARTH PLEASE POST THE END-OF-CHAPTER ANSWERS OF PHYSICS COURSE BOOK?
I AM FINDING THEM FROM YEARS AND WITH NO LUCK! IF ANY ONE HAS TEACHER'S C-D ROM, HE MIGHT HAVE ACCESS TO THAT- AS FAR AS I HAVE HEARD.

yeah i also want that.

 

[champ-student]

i

yeah i also want that.

have da cd but there are'nt any anxerzzz to them....it only have diagnosis tests.......nd some tips

 

[raamish]

i
have da cd but there are'nt any anxerzzz to them....it only have diagnosis tests.......nd some tips

maybe u are referring to the cd that comes with the book for free. The cd i was looking for doesnt come with the book called teacher's resource cd-rom.

 

[champ-student]

btw whic

maybe u are referring to the cd that comes with the book for free. The cd i was looking for doesnt come with the book called teacher's resource cd-rom.

btw which physics buk are u having??????
i'll seaarch for cd if possible

 

[champ-student]

HEY!!!!
i m having COMPLETE PHYSICS FOR O LEVEL IGCSE nd ans are given at the back.....

 

[PinkRhinos]

Can anyone help me here,a good eplanation would be of high appreciation!

"A boy is throwing a ball over a wall. He is standing 5m away from the wall,whose top is 3m highr thzn his hand at the instant he releases the ball." "Show that,if the ball is thrown with a velocity of 11m/s at an angle of 45° to the horizontal,the ball will just go over the wall.

 

[Silent Hunter]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf

6(c) and 4(a)(b) ??

thanks alot

 

[Asheeta]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

6(c) and 4(a)(b) ??

thanks alot

so for no. 4
(a) its brittle as such materials break directly
(b) take 2 points and find the gradient
so young modulus = 1/gradient

no.6
Using p = (v^2)/R
R is const
ratio= [(210^2)/R]/ [(230^2)/R]
so ratio = 0.834

 

[Asheeta]

Can anyone help me here,a good eplanation would be of high appreciation!

"A boy is throwing a ball over a wall. He is standing 5m away from the wall,whose top is 3m highr thzn his hand at the instant he releases the ball." "Show that,if the ball is thrown with a velocity of 11m/s at an angle of 45° to the horizontal,the ball will just go over the wall.

so max height reached by ball is when v = 0 m/s
so find the max height reached using (v^2) = (u^2) + 2as
0 = (11 sin 45)^2 + (2*-10*s)
s = 3.025 m
so ball goes just over the wall.
Hope it is clear.

 

[Silent Hunter]

so for no. 4
(a) its brittle as such materials break directly
(b) take 2 points and find the gradient
so young modulus = 1/gradient

no.6
Using p = (v^2)/R
R is const
ratio= [(210^2)/R]/ [(230^2)/R]
so ratio = 0.834

thank you.... and 4(c) too... thank again ... link here : http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf

 

[Asheeta]

thank you.... and 4(c) too... thank again ... link here : http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

so max stress is 9.5*10^8
using stress = F/A
9.5*10^8 = [1.9*10^3]/A
You'll obtain the min Area of X-section as 2.0 *10^-6
therefore the max area of x-section = (3.2*10^-6) - the min area

 

[Silent Hunter]

the max area of x-section = (3.2*10^-6) - the min area

didnt understand this part

 

[PinkRhinos]

so max height reached by ball is when v = 0 m/s
so find the max height reached using (v^2) = (u^2) + 2as
0 = (11 sin 45)^2 + (2*-10*s)
s = 3.025 m
so ball goes just over the wall.
Hope it is clear.

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