physics mcq help needed! urgent!

[cashem'up]

hm....but can you please explain why you took those wires as parralel??
i have a quetion....when we talk about the kinetic energy of a electron in a electric field is there a formula stating that:
  kinetic energy=potential difference across the plates * charge of the electron.

another question which  i wanted to clear was that for example i saw a question that like a specific amount of electrons and protons were moving in oppisite directions in the same wire. they asked about the total charge on it.
so can you explain whether we subtract the charges or do we add them up....

thanks.

1. Because how else would they be transmitted.......u cant have them in series.........
2. no there is isnt a formula like that but we now V/d = F/q so if if distance is constant then VQ=F where as VQ=ma..... now a= VQ/m now since we now the acceleration of particles its easy to compare the kinetic energy or if distance or time given the velocity and ke. itself can be deduced.....
 

[da_rockstar]

q1. so u mean that when wires are put together like this we always consider them as parallel?

q2.
but can't we say this:
K.E=F*d
  F=QE    but E is V/d
 F=QV/d
 K.E hence becomes QV right?
because i read in mark scheme that changing the distance of electric field strength does not effect the speed of the particle.
an what about my last one...can u help me in it?

thanks

[halosh92]

Q) two spheres A and B approach each other along the same straight line with speeds UA and UB.
the spheres collide and move off with speeds VA and VB, both in the same direction as the intial direction of sphere A.
which equation applies to an elastic collision?
1) UA  + UB =VB-VA
2)UA - UB = VB-VA
3)UA-UB=VB + VA
4) UA + UB = VB +VA

someone plzz explain
thxx

[ruby92]

Q) two spheres A and B approach each other along the same straight line with speeds UA and UB.
the spheres collide and move off with speeds VA and VB, both in the same direction as the intial direction of sphere A.
which equation applies to an elastic collision?
1) UA  + UB =VB-VA
2)UA - UB = VB-VA
3)UA-UB=VB + VA
4) UA + UB = VB +VA

someone plzz explain
thxx

----->uA           UB<------         before collision
after collison
VA------->   VB------->

taking ---> as positive
UA-(-UB)=VB-VA
UA+UB=VB-VA

[halosh92]

----->uA           UB<------         before collision
after collison
VA------->   VB------->

taking ---> as positive
UA-(-UB)=VB-VA
UA+UB=VB-VA

fankyou

[ruby92]

M/j 2007 27??

[shamanking]

can any1 help me oct/nov 09 question 9-15 and 22...i knw a lot of questions but help meeeee..p11

[halosh92]

M/j 2007 27??

A
well because first of all u should consider the direction of the stationary wave.
-------> and <---------
and this applies only to A
also consider the amplitude at first long arror, which mean high amplitude which means its an antinode....then short arrows, mean short amplitude, thats a node ull find another pair also if u look carefully
thats it.

[Phosu]

Nov 05 24

[cashem'up]

q1. so u mean that when wires are put together like this we always consider them as parallel?

q2.
but can't we say this:
K.E=F*d
  F=QE    but E is V/d
 F=QV/d
 K.E hence becomes QV right?
because i read in mark scheme that changing the distance of electric field strength does not effect the speed of the particle.
an what about my last one...can u help me in it?

thanks

well yea ur method is also rite in physics u can dervie everything from everything lol
yea for the next part i dont know the question but if i an not wrong to find the total charge u have to add them
q of proton + q of electron
electron will be negative so
Q1+(-Q2)
so Q1-Q2
in a way ur subtracting them but i think u got my point

[ruby92]

A
well because first of all u should consider the direction of the stationary wave.
-------> and <---------
and this applies only to A
also consider the amplitude at first long arror, which mean high amplitude which means its an antinode....then short arrows, mean short amplitude, thats a node ull find another pair also if u look carefully
thats it.

tnkxxx

[hesho21]

Guys plz I need help in No13 plz in Nov09,qp11 attached here

[halosh92]

Guys plz I need help in No13 plz in Nov09,qp11 attached here

which question?

[halosh92]

2003 june
Q5
Q8
someone plz explain
thx

[ruby92]

M/j 2006
22