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Physics: Post your doubts here!

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this is very easy if you were able to do part a , in part a you re asked to calculate the gradient!
you found the gradient =s/t^2 . think of this kinematic equation s=0.5 at^2 . we do not consider s= ut+0.5 a^2 because the car does not have any constant initial velocity before its acceleration .
now to calculate acceleration , a= 2 x s/t^2 that is 2 x the gradient you found . ( i am taking s as the distance here)
 
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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
 
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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.
 
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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.
 
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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)
 
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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! :)
 
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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 :)
 
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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?
 
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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. :(
 
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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.
 
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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
 
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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.
 
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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.
 
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