some questions from trial/past papers (1 Viewer)

[1234567]

2001 hsc past paper, multiple choice question 15
don't understand


explain why astraunats in orbit are weightless.


for a step up transformer, voltage is being stepped up, but accordin to the laws of conservation of momentum, the energy in must be same or larger than the output, so there must be a trade off. explain what this trade of could be


catholics 2001
question 8
in a certain experimental arrangement, the variation in the cross sectional area of a conductor on the drift velocity of the electrons was investiaged. if the original drift velocity was v ms, then after doubling the cross sectional area the new drift velocity should be
2 v, or i/2v?
the answer sais 2 v, but i thought it's 1/2v...............
coz wasn't it inversel proportional?


question 12
two rectangular coils of copper wire, each of cross sectional area 72ms are lying perperdicular to a uniform magnetic filed of flux density 2 x 10^-3 T, coil x consists of 200 turns of wire and coil y consists of 400 turns of wire, the ratio of the magnetic flux threading through coil x to that coil y is:
a 0.5
b 1.0
c. 1.5
d. 2.0

i dunt get it, please explain

 

[spice girl]

Astronauts are "weightless", because they don't experience a reaction force from their surroundings. The spacecraft is accelerating at the same rate as the astronaut, so there are no mutual forces between the two.
Like, when you read this, you feel a weight because you're sitting on a chair, and the chair is pushing up against you, in resistance to the gravitational acceleration of 9.8ms^-2 that you should be falling at. Whereas in space, both the spacecraft and the astronaut are in orbit around the earth, "falling" to the centripetal force that keeps their movement in circular orbit. That's why they're not losing altitude, yet they are "falling".

For the transformer, P = VI, and E = P*t
Since energy can never be created or destroyed, in ideal conditions E(in) = E(out).
Thus V(in)I(in) = V(out)I(out)
In reality, there are energy losses as heat through eddy currents, so E(in) >= E(out)
i.e. an inversely-proportionate decrease in current is the trade-off to an increase in voltage.

Now, let me say that I think the Catholic paper is F***ed. If the answer to Q8 is 2v, then it's wrong. Besides, it's a poorly-set question because it didn't say whether the electricity source was voltage-constant (as in batteries), current-constant (as in transistors), or neither.

Q12: You can consider coil Y as two coil X's in series. So Coil Y would experience twice the flux experienced in coil X.

 

[Jellymonsta]

some1 correct me if any of these r wrong, i am not a physics guru .

Weight is force
so f=ma
but a=0 (no gravity)
therefore f=o ie weightless.

conservation of momentum for transformers? eh? according to conservation of energy, energy out cannot be more than energy in... so for P=VI, I will decrease when V increases; energy loss also occurs, so energy out is less than energy in...

for drift velocity: I=neav
v=I/nea
if area is doubled,
v=I/ne2a
so 2v = I/neA
but that feels kinda dodgy to me...

last q... err doesnt flux=BAcos@
therefore the number of coils is irrelevant? and ratio is 1:1?

edit: i think im gonna believe spice girl over me...

 

[1234567]

for the last one,, taht ratio question
int hat case shouldn't it be 2: 1?
why is it 2:0?

 

[Jellymonsta]

its 2:1
and the others are like 3:2 and so on...
ratio is dividing one number by another; they simplified it.
i think.

 

[1234567]

stupidity of catholic paper....

 

[superhubert]

Weight is force
so f=ma
but a=0 (no gravity)
therefore f=o ie weightless.

there is acceleration ,because any direction change constitutes acceleration. The formula your looking for is the one for centrapedal acceleration a=v^2/r
they're weighless because they're in a frame of reference that is constantly falling at g, thus they are falling with the ship and experience 'weightlessness'

 

[kaseita]

Originally posted by 1234567
for the last one,, taht ratio question
int hat case shouldn't it be 2: 1?
why is it 2:0?

I don't get what your saying there, cause 2:0 isn't much of a ratio.
but I swear the answer was 1:1. The number of turns simply means how many times the wires been turned around the armature, but there's still only one armature. Also the amount of flux going through an armature is based on the area of the armature, not the number of turns.
Its not two armatures (from the question) for coil Y, so its still only one coil, with double the number of turns. As the area given is the same for both armatures, therefore the flux going through them is the same.

yea, that drift velocity one is mucked up. the answer was wrong for that one. its definitely 1/2v.

 

[BlackJack]

The problem with that is the flux will be counted many times because of the windings. Both 'Area' of coil Flux are pretty arbitrary terms.
These are based on observations of torque, and with more winding, more torque.

The area of a 200 winding coils isn't really A, it's 200A because each winding covers a complete square and is effectively another one-turn coil with area A which cuts same flux again.
If you put in 200 separate 1-turn coils of wire and connect them in parallel to the circuit, while multiplying the current 200 times, (so each coil has the same current as the 200 turn coil), it'd make no difference.

 

[Dumbarse]

what happens to drift velocity as area of the coil increases!!??
increase decrease or stay the same?!
all my books say different things

one book says it decreases because more electrons can pass through
another says it increases cause more space to pass through
another says it is not effected cause the two things cancel each other out
ahhhh, what the answer!??

 

[superhubert]

it souldn't change it, area only changes torque, the length of the conductor in creases the force but by the equation I=neva length plays no part in drift velocity. on ly, current, area of conductor as viewed from a CROSS-SECTION, charge and no of electrons....right?

 

[wogboy]

It all depends on what variables remain constant and what change.

1) If you consider the current to be constant (which is more often the case, this is the one to assume), then increasing the cross sectional area will DECREASE the drift velocity because they're inversely proportional with each other.
(V = I/nea)

2) If you consider the applied voltage to be constant (less often the case, don't assume this unless you're told), then the cross sectional area of the wire has no effect on the drift velocity and if you increase the cross sectional area of the wire, the drift velocity will remain constant. This is because as the area increases, so does the current (less resistance in the wire) so these cancel out any change in the drift velocity.

3) About the book that said that increasing the cross sectional area will INCREASE the drift velocity, this is wrong and it will never occur.