**Physics...Before/After**

MatthewRud

finality wrote: »

But didn't it say repulsive force? Gravitational force is always attractive.

But won't using Coloumb's law give you a positive answer as well? Shouldn't the answer be negative to indicate that it is a force of repulsion?

SD021

Overall I thought it was a fine paper, but very different from previous years in terms of what they gave you. My log book was open on the physical quantities page for like half the exam :P

For once I think I won't end up counting Q5 as I sorta gave sarcastic answers to the rainbow and smoke ones and left out 'due north' in the first part. Luckily it was a sort paper so I had something like 50 mins at the end left so I got another long question done and another part from Q12. Initially I thought there was a huge mistake on the paper in the nuclear question as they gave you no info. Was completely scared off it until the end but it ended up being grand. Overall might have gotten an A1 might not. If I do I'll only be a hair over 90% I'd say. I probably bled marks in all the questions.

Would love if they released some preliminary marking schemes with the papers :P

Wesc.

Liveforrugby wrote: »

40 mark window, anyone think they got the A1?

If I didn't get an A1 I would be shocked! Sooooooooo happy with how that went Turns out I didn't do great in Q5 but I did an extra question so it didn't matter

Q8: Did anyone here for (iii) look up the mass for deuterium and tritium in the log tables and multiply it by 'u'? That's what I did and in the end I got something like 3.6 MeV

Q8: for (iv) was I right in letting the charges equal 2(1.6x10^-19) and 3(1.6x10^-19) ?

Q6: for (v) did anyone get 22.3 times greater?

Q6: for (vi) how did you do that? Was it just the weight


Oh, and 11 marks for a fuse!?

mathstalk

Tankosaur wrote: »

Air pressure is the same everywhere at the same altitude, otherwise our ears would pop everytime we walked in and out of buildings.

But surely the building has central heating and hot air is less dense than cool air?

ECM2012

subz3r0 wrote: »

ECM2012 wrote: »

In the nuclear physics question, if you multiplied the mass of an electron by the mass numbers of 3H and 2H without using the values in the log tables, would you say that's ok?

No, because electrons don't come in to this question at all. They said nuclei.

You will still be given marks for the formula and substituting in and for the unit

But when calculating the difference in the mass of reactants and products you need to find the masses of the two nuclei in terms of kg..and to get that u multiply the mass number by the mass of an electron?..or no?

finality

Wesc. wrote: »

Q6: for (v) did anyone get 22.3 times greater?

Q6: for (vi) how did you do that? Was it just the weight


Oh, and 11 marks for a fuse!?

for (v) yes I got that
was vi the one where he was moving at constant velocity? Whatever they asked you to find equalled the weight, yeah :L

Bears and Vodka

ECM2012 wrote: »

But when calculating the difference in the mass of reactants and products you need to find the masses of the two nuclei in terms of kg..and to get that u multiply the mass number by the mass of an electron?..or no?

:eek:
nope. I think you're confusing this with the photoelectric effect and thermionic emission where an electron is accelerated across a cathode ray tube.

here it was all about the E=mc^2 equation.

SD021

Wesc. wrote: »

Q8: Did anyone here for (iii) look up the mass for deuterium and tritium in the log tables and multiply it by 'u'? That's what I did and in the end I got something like 3.6 MeV

Q8: for (iv) was I right in letting the charges equal 2(1.6x10^-19) and 3(1.6x10^-19) ?

Q6: for (v) did anyone get 22.3 times greater?

Q6: for (vi) how did you do that? Was it just the weight

• I got the masses of protons and neutrons and constructed the masses of the nuclei that way but it amounts to the same thing

• Yes

• No but I think I'm wrong so you're probably right :L

• Yeah forces up = forces down so yeah it's just mg.

Tankosaur

Wesc. wrote: »

Q6: for (v) did anyone get 22.3 times greater?

Q6: for (vi) how did you do that? Was it just the weight


Oh, and 11 marks for a fuse!?

6 (v) yes the answer is 22.3 = 8.5^2 / 1.8^2

yeah the answer to (vii) was just his weight which was 180kg x g ( 9 .81) which is 1760 N.

Acceleration = 0 therfore upthrust - weight = 0 weight = upthrust.

Epsi

Wesc. wrote: »

If I didn't get an A1 I would be shocked! Sooooooooo happy with how that went Turns out I didn't do great in Q5 but I did an extra question so it didn't matter

Q8: Did anyone here for (iii) look up the mass for deuterium and tritium in the log tables and multiply it by 'u'? That's what I did and in the end I got something like 3.6 MeV

Q8: for (iv) was I right in letting the charges equal 2(1.6x10^-19) and 3(1.6x10^-19) ?

Q6: for (v) did anyone get 22.3 times greater?

Q6: for (vi) how did you do that? Was it just the weight


Oh, and 11 marks for a fuse!?

1) I didn't , but I'm certain I got that question wrong.

2) I did the same , don't know if it is right.

3) Yes

4) That was the parachute opening? I did F=ma .

Wesc.

finality wrote: »

for (v) yes I got that
was vi the one where he was moving at constant velocity? Whatever they asked you to find equalled the weight, yeah :L

Yeah it was the upthrust. But like since he was moving down wouldn't the upthrust have been less than the weight? But then the force was 0 so they're equal... :S Well I just said it was the weight anyway so hopefully that's right!

Oh, thanks to everyone else too..

Tankosaur

mathstalk wrote: »

But surely the building has central heating and hot air is less dense than cool air?

At a constant pressure (charles's law) ( atmospheric pressure) hot air is less dense than cold air. Temperature has no effect on pressure.

finality

Wesc. wrote: »

Yeah it was the upthrust. But like since he was moving down wouldn't the upthrust have been less than the weight? But then the force was 0 so they're equal... :S Well I just said it was the weight anyway so hopefully that's right!

when forces are balanced an object will either be at rest or remain moving at constant velocity

Cathalog

Wesc. wrote: »

finality wrote: »

for (v) yes I got that
was vi the one where he was moving at constant velocity? Whatever they asked you to find equalled the weight, yeah :L

Yeah it was the upthrust. But like since he was moving down wouldn't the upthrust have been less than the weight? But then the force was 0 so they're equal... :S Well I just said it was the weight anyway so hopefully that's right!

Nothing to do with speed, forces are to do with acceleration.
He was moving with contant velocity, ie a=0.

Therefore, by F=ma, Fnet = 0. So forces up = forces down.

You're right, don't worry

Cathalog

finality wrote: »

Wesc. wrote: »

Yeah it was the upthrust. But like since he was moving down wouldn't the upthrust have been less than the weight? But then the force was 0 so they're equal... :S Well I just said it was the weight anyway so hopefully that's right!

when forces are balanced an object will either be at rest or remain moving at constant velocity

Newton's First Law
And ya beat me too it :P

Bears and Vodka

Wesc. wrote: »

Q8: for (iv) was I right in letting the charges equal 2(1.6x10^-19) and 3(1.6x10^-19) ?

Why though? Both deuterium and the tritium have one proton each and neutrons have a charge of 0. Why do you multiply the Mass number by the charge?

Tankosaur

My answers for q 11

(a) 8 times more ( 2^3)
(b) it says in the description not all the air can be slowed down therefore not all the energy is extracted
(c) definition of induction
(d) transformer's ( my teacher said this was right)
(e) the intensity is halfed ( decrease in level of 3b halfs the sound intensity)
(f) calculation edit: answer was 20 pi ?
(g)rectifiers ( teacher said the was also correct) diodes also work
(h) hydro energy, wave energy, geothermal etc.

Chuchoter

I have no idea what any of you are on about with electrons etc. I'd say for that question since it was a touch confusing with the not having the values like normal, will be pretty leniantly.

mathstalk

Tankosaur wrote: »

At a constant pressure (charles's law) ( atmospheric pressure) hot air is less dense than cold air. Temperature has no effect on pressure.

Remember the crushed can experiment? I just thought of the building as the can (containing the hot air). So I deduced that the outside air is pushing on the can (the building), and thus causing air from outside to enter it, bringing the smoke along with it.

Wesc.

subz3r0 wrote: »

Why though? Both deuterium and the tritium have one proton each and neutrons have a charge of 0. Why do you multiply the Mass number by the charge?

Erm, I don't know Even if that was correct when I multiplied 2 by 3 I said the answer was 5 hahaha :P But yeah I'd say I got it wrong anyway, sure it'd only be I'd say 3 marks gone

clio94

How did you find the average velocity ? Got completely mixed up for this part .. The rest of the question was fine thou.

mathstalk

clio94 wrote: »

How did you find the average velocity ? Got completely mixed up for this part .. The rest of the question was fine thou.

Distance travelled over time taken.

Cathalog

clio94 wrote: »

How did you find the average velocity ? Got completely mixed up for this part .. The rest of the question was fine thou.

Displacement divided by time. It's actually so easy, and I was doing confusing stuff to find it out. Took me like 5 minutes to figure it out. Thank fúck for Dad's Silly Triangle!!

Tankosaur

clio94 wrote: »

How did you find the average velocity ? Got completely mixed up for this part .. The rest of the question was fine thou.

The pressure decreases due to escaping air, when the lid is off hot air rises leaving a volume of x in the can, when the air cools the particles that make up x remain the same but the volume decreases, therefore the pressure decreases crushing the can

mathstalk wrote: »

Remember the crushed can experiment? I just thought of the building as the can (containing the hot air). So I deduced that the outside air is pushing on the can (the building), and thus causing air from outside to enter it, bringing the smoke along with it.

Average speed is just distance / time taken

Wesc.

clio94 wrote: »

How did you find the average velocity ? Got completely mixed up for this part .. The rest of the question was fine thou.

This is how I did it.
First I got the initial speed by using v=u+at where u=0, a= 9.7552 (I think?) and t=13
Then I subtracted the total distance from 5000 and my answer for (iii)
Then I used s=ut + 1/2at^2 to get a (wasn't sure on this)
Then I used v=u +at again to get final speed and then added that to initial and divided by 2

mathstalk

Tankosaur wrote: »

The pressure decreases due to escaping air, when the lid is off hot air rises leaving a volume of x in the can, when the air cools the particles that make up x remain the same but the volume decreases, therefore the pressure decreases crushing the can

God, why couldn't you have been my Physics teacher! Hopefully I'll get some attempt marks

clio94

Cathalog wrote: »

Displacement divided by time. It's actually so easy, and I was doing confusing stuff to find it out. Took me like 5 minutes to figure it out. Thank fúck for Dad's Silly Triangle!!

Well that makes its pretty easy .. Applied maths brain making everything over complicated ! Il get marks for my theory :P

Tankosaur

Wesc. wrote: »

This is how I did it.
First I got the initial speed by using v=u+at where u=0, a= 9.7552 (I think?) and t=13
Then I subtracted the total distance from 5000 and my answer for (iii)
Then I used s=ut + 1/2at^2 to get a (wasn't sure on this)
Then I used v=u +at again to get final speed and then added that to initial and divided by 2

The question said after the first 13 seconds " he fell 5km in the next 4 minutes and 36 seconds"

To find average speed you just divide distance by time

5000/2760 = 18.1 metres

BL1993

Cathalog wrote: »

Displacement divided by time. It's actually so easy, and I was doing confusing stuff to find it out. Took me like 5 minutes to figure it out. Thank fúck for Dad's Silly Triangle!!

I did (V + u)/2. :S

Wesc.

Tankosaur wrote: »

The question said after the first 13 seconds " he fell 5km in the next 4 minutes and 36 seconds"

To find average speed you just divide distance by time

5000/2760 = 18.1 metres

The question says: Only 4 minutes and 36 seconds more were needed to bring him down to 5km, where his parachute....

I assumed he was 5km above the ground?