PHYA5 ~ 20th June 2013 ~ A2 Physics

Original post by amish123

Inverse sqaure law ----> P = k/(x²) where P is the power output and x is the distance.

Let Ps be the power of the Sun and x1 be the distance to the Sun. Pq is the power of the quasar and x2 the distance to the quasar.

Then, Ps/Pq = x2²/x1².

Plug in the values and you'll find your power ratio to be 1¹² :smile:

.

Thanks mate. kept getting 10^-12 :L. My have got my fractions confused :smile:

Cheers!

Reply 442

Original post by posthumus

A little help here guys please :smile:

The answer to part b) ii 808K .... well anyway for part iii ... they add 35 to it... it just seems a bit too simple. Could someone please explain why this is :smile:

I mean the water was also 15 degrees celcius to start with, so I hope some you understand where my confusion lies. But in general that part of the question on a whole makes no sense to me :frown:

Thanks in avance

The copper ball goes into thermal equilibrium with the water so that the temperature of the copper ball and the beaker of water are the same. So after transferring the copper ball to it, the temperature of the copper ball is the same as the water which is 35 degrees Celsius. Given that it fell by 808Kelvin (as which know the temperature fall) and the change in kelvin is the same as the change in degrees. So actually you are adding the 808 to the 35 to get 843 degrees :smile:

Reply 443

Original post by amish123

Inverse sqaure law ----> P = k/(x²) where P is the power output and x is the distance.

Let Ps be the power of the Sun and x1 be the distance to the Sun. Pq is the power of the quasar and x2 the distance to the quasar.

Then, Ps/Pq = x2²/x1².

Plug in the values and you'll find your power ratio to be 1¹² :smile:

.

Sorry to bother you again, but I when I did this I got it different (which gave rise to answer of10^-12) which is definitely wrong. So could you please point out where I went wrong? . I said that P(S)= K/ d(S)^2 and so P(Q) = K/ d(Q)^2 where 's' represents the sun and 'Q' represents Quasar. Therefore would it not be that P(S)/d(S)^2 = P(Q)/ d(Q)^2? So P(Q)/ P(S) = d(Q)^2 / d(S)^2 through crossing right? Or at least that's what I am getting :s-smilie:

I am I being really stupid with my maths or something :/

Reply 444

Original post by JoshL123

The copper ball goes into thermal equilibrium with the water so that the temperature of the copper ball and the beaker of water are the same. So after transferring the copper ball to it, the temperature of the copper ball is the same as the water which is 35 degrees Celsius. Given that it fell by 808Kelvin (as which know the temperature fall) and the change in kelvin is the same as the change in degrees. So actually you are adding the 808 to the 35 to get 843 degrees :smile:

Ah I think I've got it now !!! So all that thermal energy (which is equivalent to 808K lets say) has gone to the water to heat it up & we've assumed that thermal equilibrium has been reached. So the whole thing is 35 degrees & since the copper was the hottest thing around it's basically dropped to 35 degrees :smile:

so that much energy (equivalent to 35K) still remains in the copper :biggrin:

Wow I think I understand it now, it makes sense!

Thank you ever so much !

Original post by smith50

Here is the discovery of an electron document + ms :wink:

Smith

I am so sorted for this exam now !!! Thank you :biggrin:

Reply 445

Nice to see some positivity on TSR! People sharing resources and helping each other, rather than

A) breezing in and cockily saying "Yeah, i'll get 100%, haven't even revised, I'm god's gift to Physics"
or B) saying "i'm so screwed for this exam" and doing nothing about it

Reply 446

amish123

Original post by JoshL123

Sorry to bother you again, but I when I did this I got it different (which gave rise to answer of10^-12) which is definitely wrong. So could you please point out where I went wrong? . I said that P(S)= K/ d(S)^2 and so P(Q) = K/ d(Q)^2 where 's' represents the sun and 'Q' represents Quasar. Therefore would it not be that P(S)/d(S)^2 = P(Q)/ d(Q)^2? So P(Q)/ P(S) = d(Q)^2 / d(S)^2 through crossing right? Or at least that's what I am getting :s-smilie:

I am I being really stupid with my maths or something :/

No worries :smile:

Okay so basically, the key thing here is the fact that K is the same for both the Sun and the Quasar. So P(S)= K/ d(S)^2 and P(Q) = K/ d(Q)^2 can be re-arranged to give you K= for both equations. So K=(d(S)^2)P(S) and also K=(d(Q)^2)P(Q). We've already estabilished the constant K must be the same so you can let both equations equal on another; (d(S)^2)P(S)=(d(Q)^2)P(Q). Can you work it out from here? :P Make sure you find P(S)/P(Q) and not the other way around, otherwise you'll get 10^-12 and not 10^12.

Reply 447

Original post by amish123

No worries :smile:

Okay so basically, the key thing here is the fact that K is the same for both the Sun and the Quasar. So P(S)= K/ d(S)^2 and P(Q) = K/ d(Q)^2 can be re-arranged to give you K= for both equations. So K=(d(S)^2)P(S) and also K=(d(Q)^2)P(Q). We've already estabilished the constant K must be the same so you can let both equations equal on another; (d(S)^2)P(S)=(d(Q)^2)P(Q). Can you work it out from here? :P Make sure you find P(S)/P(Q) and not the other way around, otherwise you'll get 10^-12 and not 10^12.

Cheers dude! I genuinely didn't know what I was doing wrong before :L

Reply 448

cooldudeman

18

does the latent heat get absorbed by the other object?

Posted from TSR Mobile

Reply 449

cooldudeman

18

Ok I really dont get it. If an ice cube dropped in water of high temp, the cube would gp to 0 celsius and change state. From ice's original temp to 0, energy is gained from the water. Then for the changing state, it that energy gained from the water too?

Posted from TSR Mobile

Reply 450

Original post by cooldudeman

Ok I really dont get it. If an ice cube dropped in water of high temp, the cube would gp to 0 celsius and change state. From ice's original temp to 0, energy is gained from the water. Then for the changing state, it that energy gained from the water too?

Posted from TSR Mobile

Erm I'm not quite sure what your asking exactly but if an Ice cube is dropped into some hot water then most likely... the Ice cubes temperature will rise to 0 degrees, then when it reaches here the energy that goes towards breaking down the inter-molecular forces in the ice, this heat energy does not increase the temperature (no increase in kinetic energy), the amount needed to change the waters state to liquid is known as latent heat of fusion. The energy is still there... it's just not being used as kinetic energy (which contributes to the temperature).

The total energy of the system is the total Kinetic energy + total potential energy stored withing inter-molecular bonds (latent energy if you like)

Hope that makes sense :smile:

Reply 451

xrpred

1

Anyone got any bets for what the long question will be this year (well, the one in thermal/nuclear physics that is). There have already been questions on radioactivity, nuclear and gas laws. So my bet is on a latent heat question!

Reply 452

Original post by xrpred

Anyone got any bets for what the long question will be this year (well, the one in thermal/nuclear physics that is). There have already been questions on radioactivity, nuclear and gas laws. So my bet is on a latent heat question!

Is there a six marker for both papers ?? :confused:

Reply 453

xrpred

1

Original post by posthumus

Is there a six marker for both papers ?? :confused:

Annoyingly >.< Yes

Reply 454

I don't know about the other paper, and I don't want to say "I'm willing to bet" on here in case someone comes around demanding money..

but for the Astro Option, I wouldn't be surprised if we had a question on the lifecycle of a star. As in, describe the various phases a star can go through from its birth to its death (protostar, main sequence star, red giant, supernova, planetary nebula, dwarf) --- all of this is on the syllabus, but it's never come up as a six-marker. Doesn't it seem RIPE to do so?

Reply 455

Original post by xrpred

Anyone got any bets for what the long question will be this year (well, the one in thermal/nuclear physics that is). There have already been questions on radioactivity, nuclear and gas laws. So my bet is on a latent heat question!

I just don't know whether there's enough to explain for latent heat to justify a six-mark question. Maybe?

Reply 456

sports_crazy

A space probe contains a small fission reactor, fuelled by plutonium, which is designed to produce an average of 300 W of useful power for 100 years. If the overall efficiency of the reactor is 10%, calculate the minimum mass of plutonium required.
energy released by the fission of one nucleus of Pu = 3.2 × 10–11 J
the Avogadro constant = 6.0 × 1023 mol–1

Is this part of the new specification. There are loads of legacy questions similar to these. I really hate these type of questions.

Reply 457

The H

1

Original post by xrpred

Anyone got any bets for what the long question will be this year (well, the one in thermal/nuclear physics that is). There have already been questions on radioactivity, nuclear and gas laws. So my bet is on a latent heat question!

Something to do with molecular kinetic theory? I'd say
If it's on the nuclear topic, then maybe radioactive decay

Reply 458

Original post by sports_crazy

A space probe contains a small fission reactor, fuelled by plutonium, which is designed to produce an average of 300 W of useful power for 100 years. If the overall efficiency of the reactor is 10%, calculate the minimum mass of plutonium required.
energy released by the fission of one nucleus of Pu = 3.2 × 10–11 J
the Avogadro constant = 6.0 × 1023 mol–1

Is this part of the new specification. There are loads of legacy questions similar to these. I really hate these type of questions.

This could easily be a question on the new specification, if they gave you the atomic mass number of plutonium (you need it and I think old physics exams had periodic tables as well)

I think the best way to do this is to work out how much energy 300W of useful power over 100 years actually is, multiply that by 10 (because of all the wasted power) and then work out how many plutonium nuclei that is

Reply 459