Award Winning Boss Posted April 24, 2012 Report Share Posted April 24, 2012 Apologies for the silly question but what's the difference between evaporation and boiling on the molecular level? Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted April 24, 2012 Report Share Posted April 24, 2012 The IB need you to know the following distinctions:Evaporation occurs at any temperature, not the boiling point, whereas boiling only occurs when the vapour pressure = atmospheric pressure (= boiling point).Evaporation leads to cooling. Boiling doesn't lead to cooling. 1 Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 24, 2012 Report Share Posted April 24, 2012 Okay! the first reply i understood but in the second reply can you please tell me if what assumptions i posted were correct, series parallel stuff. Could you clarify your question about the "series parallel stuff"? Secondly, great attempt but i dont understand certain things yet. If. for the voltage of the 1ohm resistor, why insert 'i' next to v and call it i volts? moving on, i +v =2 i get as its. total voltage and the components are in series, but this assumes that i is the voltage, but im dumbstruck when you use this equation for the graph! because what i understood tells me that i is voltage but then you use it as current ? please explain! thanks! The voltage across a resistor R ohms carrying current i amps is ( i x R) volts. Since in this case R = 1 ohm, when the current through the resistor is i amps, the voltage across it is i volts. The voltage/current characteristic shows us that when the voltage across the component X is 0.7v, the current through it is 1.3A. For this current, the voltage across the 1 ohm resistor is 1.3v. Hence the voltage across both components is (0.7 + 1.3) = 2v, the voltage supplied by the cell. Well the statements I'd written that if 'r' (internal resistance) is negligible in a parallel circuit then does it mean e= V and if r is neglegible in a series circuit does it mean e= v1 +v2 ... ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 24, 2012 Report Share Posted April 24, 2012 Well the statements I'd written that if 'r' (internal resistance) is negligible in a parallel circuit then does it mean e= V and if r is neglegible in a series circuit does it mean e= v1 +v2 ... ?Yes, if the internal resistance of a cell is negligible then its output voltage will be equal to its emf, independently of the current that it is supplying or whether it is driving a series or parallel combination of components.So if the cell were driving a number of components connected in parallel then the voltage v across each of them would be equal to the emf of the cell. e = vWhereas if the cell were driving a number of components connected in series then the emf would be equal to the sum of the voltages across each of them. e = v1 + v2 + … Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 24, 2012 Report Share Posted April 24, 2012 (edited) Hey guys, this is a bit elementary but I cannot understand the textbook. What is the Newtonian universe's basic properties or theories etc. And what does olbers paradox basically cover? The text book uses twisted sentences and it's tough to understand what is the implication and what is the knowledge we're supposed to retain from a page of nonsense! I hope this isn't a bother! Additionally the Exam questions pose a dilemma as b=L/4 pi d2 uses AU for the units of d at times and at other times uses metres. What's the real deal on it's unit? If any details are required, the questions are both of this sort: 'Show that Antares is 6.5 x 10^4 times more luminous than the sun' and the other is 'show that luminosity of Becrux is 3.43 x 10^ 3 L sun' At least to me that seems like the same thing. It hopefully is, however my second question arises here. Why is the working slightly different ? the former is : LAnt/ L sun = [bAnt x (dAnt)^2] / [bSun x (dSun)^2] ---->LAnt x L sun = [bAnt x (dAnt)^2] The latter however is the same initial formula of course but it is rearranged as : LAnt = [bAnt x (dAnt)^2] L sun You can understand my amazement I hope ! Edited April 25, 2012 by DropBoite Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted April 25, 2012 Report Share Posted April 25, 2012 Hey guys, this is a bit elementary but I cannot understand the textbook. What is the Newtonian universe's basic properties or theories etc. And what does olbers paradox basically cover? The text book uses twisted sentences and it's tough to understand what is the implication and what is the knowledge we're supposed to retain from a page of nonsense! I hope this isn't a bother! Newton’s model states that the universe is infinite, static, and uniform. This means that stars are uniformly spaced, and that if it is infinite there must be a star at every point in space (a star along every line of sight). Since there are regions without stars, Newton’s model must be inadequate. What Olber suggested contradicts what Newton says, this is where the term "paradox" come from, he argues: If stars are uniformly distributed, the number of stars giving out their light on Earth increases with the square of the distance from Earth; so number of stars is proportional to R^2. But intensity of illumination is indirectly proportional to R^2. So everywhere in the universe must be equally bright. Which is not the case, so Newton must be wrong. Additionally the Exam questions pose a dilemma as b=L/4 pi d2 uses AU for the units of d at times and at other times uses metres. What's the real deal on it's unit? If any details are required, the questions are both of this sort: 'Show that Antares is 6.5 x 10^4 times more luminous than the sun' and the other is 'show that luminosity of Becrux is 3.43 x 10^ 3 L sun' At least to me that seems like the same thing. It hopefully is, however my second question arises here. Why is the working slightly different ? the former is : LAnt/ L sun = [bAnt x (dAnt)^2] / [bSun x (dSun)^2] ---->LAnt x L sun = [bAnt x (dAnt)^2] The latter however is the same initial formula of course but it is rearranged as : LAnt = [bAnt x (dAnt)^2] L sun You can understand my amazement I hope ! Usually when calculating ratios, SI units doesn't matter. Because they end up being cancelled. So the key when calculating ratios is to make sure that you are working with consistent units. In other words, if you use the distance in AU, use it for both stars. At the end of the day you will end up with the same answer regardless of the unit used. As for your question, I will show you how to solve it, and hopefully get rid of your confusion. There is some important facts that you didn't include, so I will post the full question: For the distance, you use m - M = 5 log(d/10) 1.1-(-5.3) = 5 log(d/10) Use your calculator to find d Which happens to be = 190.5, but remember, this equation gives you the distance in parsecs. Use the data given in your data booklet (Page 3), to convert it to AU as required by the question. For the second part of the question, Bsun = Lsun/(4*pi*(Dsun)2) Bant = Lant/(4*pi*(Dant)2) So, (Bant/Bsun) = (Lant/Lsun) * (Dsun/Dant)2 (Lant/Lsun) = (Bant/Bsun) / (Dsun/Dant)2 = (4.3*10^-11) / (1/3.9*10^7)^2 = 6.5*10^4 (as required) 2 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted April 25, 2012 Report Share Posted April 25, 2012 I am want to ask who tried to Study from the IB Physics Study guide or the OSC Physics Book which one is better ? Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 25, 2012 Report Share Posted April 25, 2012 (edited) Good answer( i repped you) but that didn't answer why in the other question Lsun was multiplied will Lant for that question. Why the difference in arrangement? I am refering to the question that involves Becrux. I'm sorry I don't have the qp since I got it from the question bank the IBO sold me for an exorbitant amount The second question specifically takes the effort to convert into metres, is there some I've missed ? Thank you for your effort! *bows* Edited April 25, 2012 by DropBoite Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 25, 2012 Report Share Posted April 25, 2012 At least to me that seems like the same thing. It hopefully is, however my second question arises here. Why is the working slightly different ? the former is : LAnt/ L sun = [bAnt x (dAnt)^2] / [bSun x (dSun)^2] ---->LAnt x L sun = [bAnt x (dAnt)^2] The latter however is the same initial formula of course but it is rearranged as : LAnt = [bAnt x (dAnt)^2] L sun You can understand my amazement I hope ! Yes, I can understand your amazement! Lant/Lsun is correct. Lant x Lsun is an error. Perhaps they meant to write Lant x (Lsun)ˆ-1. 1 Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 25, 2012 Report Share Posted April 25, 2012 I didn't know that, because when I click on Reputation I get the message "Your member group is not allowed to view received reputation".timtamboy63 has now fixed this problem, so I can see reps.Thanks! Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 25, 2012 Report Share Posted April 25, 2012 I'm stumped, how do I find period/ T when I have a cosine graph, from which I can derive amplitude and wavelength? No other data are given! I think it's related to amp somehow... Graph attached. Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 25, 2012 Report Share Posted April 25, 2012 The period of the oscillation is simply the time between a pair of successive positive peaks of the waveform. It is not related to the amplitude. 1 Reply Link to post Share on other sites More sharing options...
MitchCampbell Posted April 25, 2012 Report Share Posted April 25, 2012 HIggshunter explained it very well. Keep in mind that is time is on the x-axis (which in your case it is), then the 'distance' between peaks or troughs will give you the period (you could then find the frequency, since f = 1 / period).However, if you had a distance on the x-axis (for example, position), then the 'distance' between peaks or troughs would give you the wavelength (you could use the wave equation v = f x lambda to then find the frequency or the speed of the wave, depending on what you're given) 2 Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 25, 2012 Report Share Posted April 25, 2012 The period of the oscillation is simply the time between a pair of successive positive peaks of the waveform. It is not related to the amplitude. Great! You seem to know everything in physics! Thanks Mitch for the added info, pretty useful! So, Higgs, I tried the method and got the difference between the peak at x=o and x= what seems to be 0.1125 as 0.1125 x 10^-13 s or 1.125 x 10^-14. However this is not the expected answer for T which really is: 1.1 x 10^-12 s. I'd appreciate a lead on getting to this answer Reply Link to post Share on other sites More sharing options...
Award Winning Boss Posted April 25, 2012 Report Share Posted April 25, 2012 xxI have two questions. 1) How can I improve my comprehension of the longer questions (i.e calculations etc) and do them quicker with more accuracy? I did a fair bit of practice but my physics test made me hate physics.2) Do you like the colour pink? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 25, 2012 Report Share Posted April 25, 2012 The period of the oscillation is simply the time between a pair of successive positive peaks of the waveform. It is not related to the amplitude. So, Higgs, I tried the method and got the difference between the peak at x=o and x= what seems to be 0.1125 as 0.1125 x 10^-13 s or 1.125 x 10^-14. However this is not the expected answer for T which really is: 1.1 x 10^-12 s. I'd appreciate a lead on getting to this answer The time between the first two peaks on the graph appears to be slightly longer than that between the subsequent ones, which is exactly 22 x 0.005 x 10ˆ-13 s = 1.1 x 10ˆ-14 s, not the expected 1.1 x 10ˆ-12 s. Can you give more details of the question? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 25, 2012 Report Share Posted April 25, 2012 1) How can I improve my comprehension of the longer questions (i.e calculations etc) and do them quicker with more accuracy? I did a fair bit of practice but my physics test made me hate physics.Many of the questions are derived from different perspectives on the same theory. So going through past exam papers should be a fine way of becoming familiar with the calculations and getting up to speed.2) Do you like the colour pink?Yes, my wife looks great in pink! Reply Link to post Share on other sites More sharing options...
Award Winning Boss Posted April 25, 2012 Report Share Posted April 25, 2012 Many of the questions are derived from different perspectives on the same theory. So going through past exam papers should be a fine way of becoming familiar with the calculations and getting up to speed. Thanks, I'll be sure to keep on practising. But I don't want to do all the past paper questions ... unless I try higher questions Yes, my wife looks great in pink! Any specific shade? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 25, 2012 Report Share Posted April 25, 2012 Yes, my wife looks great in pink!Any specific shade?The one in your avatar looks just fine!Or were you referring to the pink/mauve colour that the admins have chosen for the new group in which they have put Mitch and I? Reply Link to post Share on other sites More sharing options...
Award Winning Boss Posted April 25, 2012 Report Share Posted April 25, 2012 The one in your avatar looks just fine! Or were you referring to the pink/mauve colour that the admins have chosen for the new group in which they have put Mitch and I? That's a nice colour. I recommend she tries a really fiery red. To spice the wardrobe up a little bit. Reply Link to post Share on other sites More sharing options...
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