Eydie Posted January 5, 2011 Report Share Posted January 5, 2011 Thank you very much people!More questions? [HL, Gravitation and Orbital Motion] A satellite is orbiting the earth in a circular orbit. Which one of the following properties of the satellite does not remain constant?A. Kinetic energyB. Gravitational potential energyC. Angular momentumD. VelocityI think A and B should be both constant since the R is not changing. So it's either C or D. What is angular momentum, though? How is it different from momentum? I think velocity should be constant but may be not due to acceleration and air resistance... (even though I think air resistance is negligible)I'm not quite sure what angular momentum is either... But I think velocity shouldn't be constant since velocity is a vector quantity and a satellite changes direction all the time... Thus even if it's going at the same speed it's velocity is not constant due to constant change of direction, perhaps? 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted January 5, 2011 Report Share Posted January 5, 2011 I'm not quite sure what angular momentum is either... But I think velocity shouldn't be constant since velocity is a vector quantity and a satellite changes direction all the time... Thus even if it's going at the same speed it's velocity is not constant due to constant change of direction, perhaps?OHHH haha yeah I forgot thank you! So velocity changes with orbital position.. but if we talk about speed, it remains constant right? I get that now, thank you! Btw I finished my MCQ holiday homework so you guys should be happy as I won't ask more questions soon (I may need help with lab report later but that's another business) so thank you guys very much for helping me! Reply Link to post Share on other sites More sharing options...
saunders is god Posted January 24, 2011 Report Share Posted January 24, 2011 Any physics students out there please help!!!A lead bullet is fired into an iron plate where is deforms and stops. As a result, the temperature of the lead increases by an amount ΔT. For a lead bullet having twice the mass but the same spee d of impact, what would be the best estimate of its temperature increase? A. 1/2 ΔT B. ΔT C. √2 ΔT D. 2 ΔTThanks in advance Reply Link to post Share on other sites More sharing options...
JustAnotherAsian Posted January 24, 2011 Report Share Posted January 24, 2011 is it A?I used Q=mcΔT where m = 2Q=2mcΔTΔT= Q/2mcor A Reply Link to post Share on other sites More sharing options...
genepeer Posted January 24, 2011 Report Share Posted January 24, 2011 The heat used to increase temperature comes from the kinetic energy of the bullet.Q = 1/2mv2 = mcΔTΔT = v2/2cThe temperature increase has no relation with the mass of the bullet, so B is the answer. 1 Reply Link to post Share on other sites More sharing options...
sid1729 Posted January 24, 2011 Report Share Posted January 24, 2011 D ... velocity is not constant as the direction is changing!!Angula momentum is the angle swept by the satellite in a period of time... which is bound to stay the same in space! Reply Link to post Share on other sites More sharing options...
fan Posted January 24, 2011 Report Share Posted January 24, 2011 As Gene-Peer said temperature change does not depend on the mass. So B is the answer Reply Link to post Share on other sites More sharing options...
saunders is god Posted January 30, 2011 Report Share Posted January 30, 2011 thanks guys yeah b was the answer, it was a past paper question Reply Link to post Share on other sites More sharing options...
nametaken Posted February 10, 2011 Report Share Posted February 10, 2011 I have a question which I'm struggling to get to grips with. This question is in four parts (I've done the first part, so I'll show you the other three parts):1) Monochromatic light of wavelength 5.4×10^-7 m is incident on a metal surface which has a work function of 1.40 × 10^-19 J.a) Calculate the maximum kinetic energy of an electron emitted from the surface.I was thinking to use KE= 0.5mv^2, but I don't know the velocity of the electron, although I know that the mass of one electron is 9.1 × 10^-31 kg. b) Calculate the maximum speed of the emitted electron.I have no idea how to do this.....c) Calculate the de Broglie wavelength of the fastest electrons.I know that the general equation for the de Broglie wavelength for a particle that has mass is h/√2πmKT, where m is the mass, K is the Boltzmann's constant, T is the temperature, and h is Planck's constant. I also know that I could use the mass, 9.1 × 10^-31 kg for this equation, but I don't know the temperature. Can the temperature be worked out using the work function and the wavelength? Or am I just getting everything completely wrong here? Thanks for your help!!D. Reply Link to post Share on other sites More sharing options...
fan Posted February 19, 2011 Report Share Posted February 19, 2011 (edited) I added a pdf version so that its easier to read, so you might look at that rather then this if you would likeInformation provided:λ = 5.4 x 〖10〗^(-7)m ɸ (work function) = 1.40 x 〖10〗^(-19) JEinstein’s equation to find the energy carried by a photon:E=hfElectrons absorb different amounts of photons. If the photon is large enough it will give electrons enough energy to leave the surface. This energy is known as the work function. If there is any extra energy it will be kept by the electron as kinetic energy.Hence the incoming energy of the photon will equal the energy required for the electron to leave the surface plus the kinetic energyhf= ɸ+KENow frequency can be written as:f= c/λ = (3.0 x 〖10〗^8)/(5.4 x 〖10〗^(-7) ) = …….(Sorry don’t have a calculator at the moment . so you will have to do the calculating)Now your question says to find the KE, so let’s re-arrange the equation:KE= hf- ɸh = plank constant = 6.63 x 〖10〗^(-34) Therefore KE = ((6.63 x 〖10〗^(-34)) x f) - (1.40 x 〖10〗^(-19))b) Now the next question is asking about the speed KE = 1/2 mv^2Make v the subject V = √ 2KE/mm = mass of an electronc) De Broglie’s equation is λ= h/pp being momentumNow KE = 1/2 mv^2 and p=mvv = p/msubstitute in the KE equation and we getKE = p^2/2mMake p the subjectp = √(KE x 2m)After getting p we can find the De Broglie wavelength λ= h/pquestion.pdf Edited February 19, 2011 by ∑fan 1 Reply Link to post Share on other sites More sharing options...
flamicecream Posted February 22, 2011 Report Share Posted February 22, 2011 From the May 08 SL paper 2:The initial volume of the gas is 0.20 m3 at a temperature of 27C and a pressure of 100,000 Pa. After compression the volume of the gas is 0.070 m3 and the temperature is 57 C. Determine the final pressure of the gas.I know it has something to do with PV=nRT but I'm not sure what to do.. Reply Link to post Share on other sites More sharing options...
fan Posted February 22, 2011 Report Share Posted February 22, 2011 well you have v= 0.20 m3 T= 27C P1 =100,000 Paafter compression you get v= 0.070 m3 T=57 Cso what we do first is find a value that will be the same in both of them. This will be the number of moles.PV=nRTPV/RT = nR is the gas constant(100,000 x 0.20)/ (8.31 x 27) = 89.14 molesso now we have v= 0.070 m3 T=57 C R=8.31n =89.14 molessubstitute the values intoP2=nRT/v and we get the the final pressure final pressure is 603,174 Pa 1 Reply Link to post Share on other sites More sharing options...
flamicecream Posted February 23, 2011 Report Share Posted February 23, 2011 well you have v= 0.20 m3 T= 27C P1 =100,000 Paafter compression you get v= 0.070 m3 T=57 Cso what we do first is find a value that will be the same in both of them. This will be the number of moles.PV=nRTPV/RT = nR is the gas constant(100,000 x 0.20)/ (8.31 x 27) = 89.14 molesso now we have v= 0.070 m3 T=57 C R=8.31n =89.14 molessubstitute the values intoP2=nRT/v and we get the the final pressure final pressure is 603,174 PaThanks, that's roughly what I got Reply Link to post Share on other sites More sharing options...
flamicecream Posted February 26, 2011 Report Share Posted February 26, 2011 Another question: I took data measuring the period of a pendulum. I set my uncertainty for my measurements as +/- 0.15 seconds, and took 5 trials. What is the uncertainty of the average period? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted February 27, 2011 Author Report Share Posted February 27, 2011 5(0.15)=0.75s Reply Link to post Share on other sites More sharing options...
fan Posted March 1, 2011 Report Share Posted March 1, 2011 Alex aren't you meant to find the percentage uncertainty of each value and then add them?e.gTime (s) ± 0.051.002.00(0.05/1.00) x 100 = 5%(0.05/2.00) x 100 = 2.5%total percentage uncertainty = 7.5%average time = (1.00 + 2.00)/2 = 1.5then you find 7.5% of 1.5 and then you get your uncertainty. Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 2, 2011 Author Report Share Posted March 2, 2011 That would make much more sense than mine I really don't do much with uncertainties, just include them and explain how I got it Flamicecream, do the % thingy Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 23, 2011 Author Report Share Posted March 23, 2011 (edited) So, my IA is currently for option G (electromagnetic waves) and I decided to "prove" snell's law. I found a glass rectangular prism and a laser for this and all that good stuff but was wondering if anyone could verify that my thought process behind the math isn't out of whack...K, so snell's law is nsin(theta1)=nsin(theta2)For my diagram theta 2 is being used so I'm going to call it theta3 for snell's law so nsin(theta1)=nsin(theta3)theta3=theta1+theta2The way I planned on getting theta2 was law of sines (sin a)/A=(sin b)/B but using a drawn in triangle (as you can see in the diagram) so (sin90)/h=(sin theta2)/xSo theta2=sin-1[x(sin90)]/hAll in all I got snell's law down to nsin(theta1)=nsin[sin-1(xsin(90)/h)+theta1]This all make sense or did I screw it all up? Edited March 25, 2011 by Drake Glau Reply Link to post Share on other sites More sharing options...
Rigel Posted April 7, 2011 Report Share Posted April 7, 2011 I need help on solving this:This question is about Kinematics.Someone is walking in the street, when suddenly he sees that a woman is falling down of the last floor of a building, so he uses his superpowers and starts flying with a constant acceleration of 15m/s2. The woman is falling down from a distance of 300m. Then,a) What would be the length that the woman fell down up to when she was rescued by the man?b) How much will the man take to save the woman?c) What would be the speed of the man and the woman when the man rescues the woman? Thank you very much for your help! I tried solving it, but i can't get the ammount of metres the woman fell down... Reply Link to post Share on other sites More sharing options...
genepeer Posted April 7, 2011 Report Share Posted April 7, 2011 Assuming he was standing directly underneath her when he started flying vertically upwards, that he started flying at the exact moment she started falling, and that acceleration by gravity is 10ms-2Man's distance from the last floor = 300 - Distance he flies= 300 - at2/2= 300 - 15t2/2Woman's distance from the last floor = 10t2/2At the point he rescues her, both distances are equal:300 - 15t2/2 = 10t2/2300 = 25t2/224 = t2t = 4.90sIt seems I had to solve b) before solving a) a) Woman's distance from last floor = 10t2/2= 10t2/2= 10*24/2= 120mb) Time taken to reach woman = 4.90sc) Speed of man = at= 15*4.9= 3010ms-1 to 3sf.This final speed is too high and would kill instead of saving the girl. Should I take into consideration the effect of gravitational pull on the man? Which would cause he's (effective) acceleration to be 5ms-2 not 15ms-2 and his speed upon rescue to be much less (31.6ms-1). Maybe this had been taken into consideration when his acceleration was given, another assumption I made to answer the problem. This is not a good problem as I had to make many assumptions... Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.