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Multiple Choice (2 Viewers)

rithvikt

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i still dont get how 1=A...velocity of the ball dosent change...horizontal velocity is constant and in one direction...verticle velocity changes direction and =0 at the max height.....and also by using this link
HTML:
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
when u put in a value for velocity... the starting velocity = final...

i always get confused with projectile....can some one please clear this up...if u have done so already can u do it again....
 

sweet_chick

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Mellonie said:
12 is not A liek eevyone keeps saying...

plank couldnt explain what he had found, einstein instead explained it.

HAlf the multiple answers everyone putting up is wrong, and does not make sense!!!! N i dont want to put mine up cuz its prolly equally wrong, but can someone put somethnig up which is coreect like form teachers. And y is ppl putting multiple choice answers up syaing this isdefinately correct making their selves look so stupid.

Planck explianed black body radiation using the idea of quanta....the question doesnt ask who explained how quanta can work....it asks who explained the relationship in the graph
 

sweet_chick

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jarro_2783 said:
there is one problem with the textbook which you failed to notice. The question in the exam said "... relationship between the intensity of black body radiation...", your textbook said "... on the photoelectric effect...". Your textbook and the question are two different things, planck is the correct answer.

These are the answers, don't argue, they're right. I'll explain each one as well.

9. C
Eddie currents will slow down the ring in Q. P is plastic and R isn't connected so no currents. Therefore C.
This isnt right. The unconnected copper ring still has eddy currents induced in it. think about it, a straight piece of metal can have eddy currents induced in it, and it isnt joined at each end. so the answer is B because it has less eddy currents induced than the closed ring.
 

Captain Gh3y

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rithvikt said:
i still dont get how 1=A...velocity of the ball dosent change...horizontal velocity is constant and in one direction...verticle velocity changes direction and =0 at the max height.....and also by using this link
HTML:
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
when u put in a value for velocity... the starting velocity = final...

i always get confused with projectile....can some one please clear this up...if u have done so already can u do it again....
Do projectiles travel in a straight line?

That's right. Therefore the velocity is constantly changing.
 

thunderdax

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OK I'll try to list possible options for each question:
1. A
2. D
3. B
4. unsure (although pretty sure its C or D)
5. C
6. B (although a wierd question cause resistance could come into it)
7. A or D
8. A or C
9. B or C
10. B or C
11. B
12. A (This is blackbody radiation, not the photoelectric effect)
13. A or C (Although im almost certain its C)
14. B
15. D

Now, for 7, ppl who answered D explain this. The coil rotates 90 degrees to the vertical. Take the bottom end. Using r.h.palm rule, this end has a force going to the left. But that is the same way it is being rotated. Since Lenz's law states the force needs to oppose the movement, how is it D?

And for 9, the copper ring on R is broken. How can there be circular currents going around it? I'm uncertain, but i still stand by C.
 

Captain Gh3y

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thunderdax said:
OK I'll try to list possible options for each question:
1. A
2. D
3. B
4. unsure (although pretty sure its C or D)
5. C
6. B (although a wierd question cause resistance could come into it)
7. A or D
8. A or C
9. B or C
10. B or C
11. B
12. A (This is blackbody radiation, not the photoelectric effect)
13. A or C (Although im almost certain its C)
14. B
15. D

Now, for 7, ppl who answered D explain this. The coil rotates 90 degrees to the vertical. Take the bottom end. Using r.h.palm rule, this end has a force going to the left. But that is the same way it is being rotated. Since Lenz's law states the force needs to oppose the movement, how is it D?

And for 9, the copper ring on R is broken. How can there be circular currents going around it? I'm uncertain, but i still stand by C.
Question 7 should be cancelled because it's such a shit diagram.
 

rithvikt

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thunderdax said:
OK I'll try to list possible options for each question:
1. A
2. D
3. B
4. unsure (although pretty sure its C or D)
5. C
6. B (although a wierd question cause resistance could come into it)
7. A or D
8. A or C
9. B or C
10. B or C
11. B
12. A (This is blackbody radiation, not the photoelectric effect)
13. A or C (Although im almost certain its C)
14. B
15. D

Now, for 7, ppl who answered D explain this. The coil rotates 90 degrees to the vertical. Take the bottom end. Using r.h.palm rule, this end has a force going to the left. But that is the same way it is being rotated. Since Lenz's law states the force needs to oppose the movement, how is it D?

And for 9, the copper ring on R is broken. How can there be circular currents going around it? I'm uncertain, but i still stand by C.
even if the ring in R is broken..it still induces a current( faraday)...
 

B.J

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thunderdax said:
OK I'll try to list possible options for each question:
1. A
2. D
3. B
4. unsure (although pretty sure its C or D)
5. C
6. B (although a wierd question cause resistance could come into it)
7. A or D
8. A or C
9. B or C
10. B or C
11. B
12. A (This is blackbody radiation, not the photoelectric effect)
13. A or C (Although im almost certain its C)
14. B
15. D

Now, for 7, ppl who answered D explain this. The coil rotates 90 degrees to the vertical. Take the bottom end. Using r.h.palm rule, this end has a force going to the left. But that is the same way it is being rotated. Since Lenz's law states the force needs to oppose the movement, how is it D?

And for 9, the copper ring on R is broken. How can there be circular currents going around it? I'm uncertain, but i still stand by C.
FOR QUESTION 7, turn the page 90 degrees and it will show the diagram like they normally show it. B and C are automatically ruled out because current doesnt flow like that. Using Right Hand Rule, D is the one that workd.
THey have forces go against eachother hence why they use split ring commutators so it doesnt stop at this angle of rotation
7 im willing to bet on is D

FOR QUESTION 9, Q has no currents thus falls straight down. R is split, it still has eddy currents because magnetic flux is changin around the metal. It just forms little circulating currents and has a little bit of resistance to falling. Q has maximum resistance because eddy currents are bigger and stronger there 9 is B. I had similar quesion in trial paper so im sure its B
 

rithvikt

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Captain Gh3y said:
Do projectiles travel in a straight line?

That's right. Therefore the velocity is constantly changing.
what do u mean!...for the projectile to be travelling in a parabolic path, the horizontal velocity needs to be a constant..dont it?..dont u know that experiment with that mass dropped from the mast of the ship when moving...if the velocity was changing then it wouldnt even have a parabolic path..
 

Ben-d0g

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You're answering your own question. Total velocity is the combination of horizontal and vertical velocity. Even though horizontal is constant, vertical is always changing, therefore velocity as a whole is changing.
 

B.J

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on second thoughts of question 7 i cant tell which one RHR works for. I put D so i hope its that but you people are making me think its A now
 

rithvikt

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Ben-d0g said:
You're answering your own question. Total velocity is the combination of horizontal and vertical velocity. Even though horizontal is constant, vertical is always changing, therefore velocity as a whole is changing.
damn questions r so tricky...yeh i kinda got what ur saying
 

thunderdax

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B.J said:
FOR QUESTION 7, turn the page 90 degrees and it will show the diagram like they normally show it. B and C are automatically ruled out because current doesnt flow like that. Using Right Hand Rule, D is the one that workd.
THey have forces go against eachother hence why they use split ring commutators so it doesnt stop at this angle of rotation
7 im willing to bet on is D

FOR QUESTION 9, Q has no currents thus falls straight down. R is split, it still has eddy currents because magnetic flux is changin around the metal. It just forms little circulating currents and has a little bit of resistance to falling. Q has maximum resistance because eddy currents are bigger and stronger there 9 is B. I had similar quesion in trial paper so im sure its B
Turning the paper 90 degrees for 7 does nothing. It just confuses things. The coil is going into the page.
 

B.J

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thunderdax said:
Turning the paper 90 degrees for 7 does nothing. It just confuses things. The coil is going into the page.
haha i dunno, helped me understand what was going on
 

minushuman

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Alrighty then.

Question 1
D, that is the correct answer. Well it's a correct answer, from what I see A, B and D could all be true, but I put D. D is indeed correct because throughout the motion the ball is ongoing constant horizontal velocity, hence there is no horizontal acceleration ever. At the highest point the ball is at equilibrium between the initial force and that of gravity, it is neither moving up or down. As such the acceleration of the ball at the top of the motion is zero.

Question 5
The answer is C. Initialy = usinx, at 40deg ~ 32ms, 50deg ~38ms, therefore it will take longer for the force of gravity to pull the 50degree launch back to ground. I answered B for this, cause I wasn't thinking so clearly :)

Hmm, it's free maccas time, i'll b e back later.
 

rithvikt

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TEACHERS results:

1.A
2.D
3.B
4.D
5.C
6.B
7.D *i think*
8.C
9.B
10.B
11.B
12.A
13.A or C *forgot*
14.B
15. A *i think*
 

lum

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Originally Posted by thunderdax
OK I'll try to list possible options for each question:
1. A
2. D
3. B
4. unsure (although pretty sure its C or D)
5. C
6. B (although a wierd question cause resistance could come into it)
7. A or D
8. A or C
9. B or C
10. B or C
11. B
12. A (This is blackbody radiation, not the photoelectric effect)
13. A or C (Although im almost certain its C)
14. B
15. D

Now, for 7, ppl who answered D explain this. The coil rotates 90 degrees to the vertical. Take the bottom end. Using r.h.palm rule, this end has a force going to the left. But that is the same way it is being rotated. Since Lenz's law states the force needs to oppose the movement, how is it D?

And for 9, the copper ring on R is broken. How can there be circular currents going around it? I'm uncertain, but i still stand by C.
B.J said:
FOR QUESTION 7, turn the page 90 degrees and it will show the diagram like they normally show it. B and C are automatically ruled out because current doesnt flow like that. Using Right Hand Rule, D is the one that workd.
THey have forces go against eachother hence why they use split ring commutators so it doesnt stop at this angle of rotation
7 im willing to bet on is D

FOR QUESTION 9, Q has no currents thus falls straight down. R is split, it still has eddy currents because magnetic flux is changin around the metal. It just forms little circulating currents and has a little bit of resistance to falling. Q has maximum resistance because eddy currents are bigger and stronger there 9 is B. I had similar quesion in trial paper so im sure its B
ok, first of all for question 7 i reckon it's A, because first of all, you CAN'T use the right hand grip rule with the diagram they provide. now imagine the loop was spun 90degrees, so the wires were at the top and bottom. NOw use the right hand grip rule, and as the current remains in the same direction, you get A. (remember it's a convertional current the diagram's showing.), but then again, i'm only 65% sure.

for 8 i hope they cancel it :|, caz i thought it was a stupid AC and put d :(

for 9., i agree wif person on top, it's B, even thought the copper is not a "complete" circuit, but even a piece of metal would have eddy currents, and how is that a complete circuit? it's B, i'm 95% sure.

for 10 i can confirm it's B, look at the jacaranda text book, 100%

for 13. i'm 85% sure it's C, as the extra electron would mean it has more valence electrons, hence the gap from conduction band would be smaller than the valence band of the silicon.


Here's a revised set of solutions.

1. A
2. D
3. B
4. C or D
5. C
6. B (60%)
7. A or D
8. A or C
9. B
10. B
11. B
12. A
13. A or C (85% C)
14. B
15. D

damn i screwed that mc hard...
 

seadonkey

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1A
2D
3B
4D
5C
6B *i could b wrong for this* in Q it says "wire of very low resistance"
7A remember its a GENERATOR
8C
9B
10B
11B
12A
13A
14B
15D use right hand palm rule... electrons up holes down
 

who_loves_maths

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Question 13

Originally Posted by thunderdax
13. C (since extra electron in the valence band. If it was A, the conduction band would be slightly lower)
Originally Posted by KFunk
13. C (well... you wouldn't *see* a hole 'level' would you?)
i put A - Boron for Q13.

the "dopant level" resembles the valence energy level of the extrinsic (doped) semiconductor. and since it is higher than the intrinsic valence level of the semiconductor, then the semiconductor must have been doped with a Group 3 element.

Reason: Along the same Period of the periodic table, Group 3 elements have a higher energy valence shell than Group 4 elements, this is because as we go across a Period from left to right, the size of the atom of the elements decrease - ie. their outermost shells' radii decrease {ppl doing Chemistry would know this trend}. This is due to increasing positive charge (protons) in the nucleus as we go across a period, resulting in the electrostatic attractive forces between the nucleus and single electrons in the outershell greater, drawing them further in toward the nucleus.

hence, Group 3 elements have large valence radii than Group 4.

then, esp. ppl who do Quanta to Quarks would know, as the radii of shells increase, so do the energy of that shell. hence, Group 3 elements have higher energy valence shells than Group 4 elements.

so, by doping a semiconductor with a Groups 3 atoms, you would raise the overall valence energy of the conductor - raising the dopant energy level.

ie. the answer is the only Group 3 element in the choices - being Boron.
 

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