Chronofox Posted May 8, 2011 Report Share Posted May 8, 2011 No?Well...How would you solve it if the mirror was concave, anyways? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted May 8, 2011 Author Report Share Posted May 8, 2011 (edited) No?Well...How would you solve it if the mirror was concave, anyways?Well, object needs to be between the focal point and the mirror.-(di/do)=2(1/do)+(1/di)=1/15So you need a di twice as big as the do and when you add their inverses you get 1/15...45di and -22.5do (negative because it's on the left side of the mirrorI seriously just did that my process of elimination of numbers divisible by 15... -(45/-22.5)=2(1/22.5)+(1/45)(2/45)+(1/45)=3/45=1/15Well, that's not between the focal point and the mirror...I don't know whats going on now =/ Edited May 8, 2011 by Drake Glau Reply Link to post Share on other sites More sharing options...
flamicecream Posted May 8, 2011 Report Share Posted May 8, 2011 Two questions: i) The solar intensity at the position of the Earth is 1380 W/m2. The average albedo of Earth is 0.300. Show that the average reflected intensity from the Earth is about 100 W/m2. ii) One of the expected results of global warming is an increased sea level. The increase in volume ΔV for a temperature ΔT is given by ΔV=ϒVΔT. Show, using the following data, that the resulting rise in sea level is about 0.5m. Temperature increase = 2.0 CSurface area of oceans on Earth= 3.6E8 km2Average ocean depth = 3.0 kmϒ = 8.8E-5 Reply Link to post Share on other sites More sharing options...
Drake Glau Posted May 9, 2011 Author Report Share Posted May 9, 2011 Two questions: i) The solar intensity at the position of the Earth is 1380 W/m2. The average albedo of Earth is 0.300. Show that the average reflected intensity from the Earth is about 100 W/m2. ii) One of the expected results of global warming is an increased sea level. The increase in volume ΔV for a temperature ΔT is given by ΔV=ϒVΔT. Show, using the following data, that the resulting rise in sea level is about 0.5m. Temperature increase = 2.0 CSurface area of oceans on Earth= 3.6E8 km2Average ocean depth = 3.0 kmϒ = 8.8E-5Umm I'm not sure on the the first one, albedo is the reflecting coefficient so I would think that'd be 1380*0.300 but that equals 414, not 100. Sorry Use the equation and plug stuff in. Your initial volume will be the surface area*dept.(8.8E-5)*(3.6E8*3)*2=deltaVI came up with 190,080km^3But that was with a crappy computer calculator Reply Link to post Share on other sites More sharing options...
Req Posted May 11, 2011 Report Share Posted May 11, 2011 v = atd=vt+1/2at², the v is initial velocity which is 0 so...d = vt = 1/2 at²d = 1/2 a(d/v)² = 1/2 a d²/v²a = 2v²/dF=maF=m(2v²/d)F/m=2v²/dFd/m=2v²Dividing by two, multiply by 0.5, same thing so(1/2)Fd/m is the same as Fd/2m soFd/2m=v² thereforev=sqrt[Fd/2m]Friction and the force you're pushing with work in opposite directions ( ) so your F=Fp-FfAnser should be D. v=sqrt[(Fp-Ff)d/2m]Sorry I believe it's C as we use work-energy theorem:F.x=(mv^2)/2Given that F=Fp-Ff and x=d we substitute and solve for v...(Fp-Ff)2d=mv^2v=sqrt[(Fp-Ff)2d/m]----> answer D Reply Link to post Share on other sites More sharing options...
Req Posted May 11, 2011 Report Share Posted May 11, 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 advanceIt's D. Given that Temperature is proportional to the avg Kin Energy, we know that KE=mv^2/2, if mass is doubled then the temp is doubled--->2 ΔT Reply Link to post Share on other sites More sharing options...
genepeer Posted May 12, 2011 Report Share Posted May 12, 2011 Sorry I believe it's C as we use work-energy theorem:F.x=(mv^2)/2Given that F=Fp-Ff and x=d we substitute and solve for v...(Fp-Ff)2d=mv^2v=sqrt[(Fp-Ff)2d/m]----> answer DYou just contradicted yourself and confirmed Drake's solution in fewer lines! 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 advanceIt's D. Given that Temperature is proportional to the avg Kin Energy, we know that KE=mv^2/2, if mass is doubled then the temp is doubled--->2 ΔTKE = mv^2/2 is the total kinetic energy.Average kinetic energy would be mv^2/(2n), n being the number of particles in the bullet. Doubling the mass also doubles the number of molecules, the two cancel and the average kinetic energy stays the same. --> ΔT (B) Reply Link to post Share on other sites More sharing options...
dessskris Posted May 15, 2011 Report Share Posted May 15, 2011 I know this is very stupid, but anyway I need to know how to do this type of question. For example:A car drives the first 10 m with a constant speed for 3 s, the next 5 m with a different constant speed for 4 s and the last 7 m with yet a different constant speed for 6 s.How do I find the average speed?I usually do it by:Speed = (10*3 + 5*4 +7*6)/(3+4+6) = 7.1 m/sI think there should be a better and simpler method than that.How would you guys do this question? Is there any fixed formula for this type of question?Would it be easier if there was a distance/time graph? How to do it then? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted May 15, 2011 Author Report Share Posted May 15, 2011 (edited) I know this is very stupid, but anyway I need to know how to do this type of question. For example:A car drives the first 10 m with a constant speed for 3 s, the next 5 m with a different constant speed for 4 s and the last 7 m with yet a different constant speed for 6 s.How do I find the average speed?I usually do it by:Speed = (10*3 + 5*4 +7*6)/(3+4+6) = 7.1 m/sI think there should be a better and simpler method than that.How would you guys do this question? Is there any fixed formula for this type of question?Would it be easier if there was a distance/time graph? How to do it then?[(10/3)+(5/4)+(7/6)]/3Find the speeds then average them is what I'd do. 1.92m/s is what I get.I'm not sure why you are multiplying your distance by your time. It isn't traveling 10m in 1 second, it's traveling 10m in 3. Are you just adding the distance total and then dividing by the total time?10+5+6/(3+4+6)21/13=1.62m/sOr I guess you could do a weighted average and use the time/total time to find your percents that each speed was occurring during the total time.(10/3)*(3/13)=0.77(5/4)*(4/13)=0.38(7/6)*(6/13)=0.54Total average speed equaling 1.69I don't like how I got 3 different numbers but I would personally go with the weighted average one because it takes into account how long a speed lasted where a normal average wouldn't. Edited May 15, 2011 by Drake Glau 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted May 15, 2011 Report Share Posted May 15, 2011 oh geez what am I doing!mmm I think it should be (10+5+7)/(3+4+6) then? so 1.69.I think your weighted average and percentage thing makes the most sense. I'll go with that one. thank you!but I'm still waiting for answers from other people because I really want to be sure because I don't want to get this kind of question wrong! it's quite an easy mark and I don't wanna lose it... thanks! Reply Link to post Share on other sites More sharing options...
Rigel Posted May 15, 2011 Report Share Posted May 15, 2011 oh geez what am I doing!mmm I think it should be (10+5+7)/(3+4+6) then? so 1.69.I think your weighted average and percentage thing makes the most sense. I'll go with that one. thank you!but I'm still waiting for answers from other people because I really want to be sure because I don't want to get this kind of question wrong! it's quite an easy mark and I don't wanna lose it... thanks!I'd go with Drake's answer of 1.69. When those kind of problems appear, my method is to sum the total distance and the total time, and divide those values. Bam! You will get the average velocity/speed that way. 2 Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted May 26, 2011 Report Share Posted May 26, 2011 I'm pretty sure I'm missing something really obvious Reply Link to post Share on other sites More sharing options...
nametaken Posted May 26, 2011 Report Share Posted May 26, 2011 (edited) Well, the formula for momentum is mass * velocity.And the formula for Kinetic energy is 1/2mv2.......... Edited May 26, 2011 by nαmetαken Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted May 26, 2011 Report Share Posted May 26, 2011 Yeah, I tried it, and I got v1/v2 which isn't an answer Reply Link to post Share on other sites More sharing options...
Sublime Sunshine Posted May 26, 2011 Report Share Posted May 26, 2011 (edited) You are trying to get the kinetic energy formula purely in terms of the masses, therefore, you replace velocity with momentum: K1 = (1/2)x(m1)x(v1)2 divided by K2 = (1/2)x(m2)x(v2)2 The (1/2)'s cancel, and in the next step, you use the formula p = mv and replace v in the equation with (p/m). (m1)x(p1/m1)2 divided by (m2)x(p2/m2)2 One of the masses in each equation cancels, leaving one mass on the bottom and the momentum squared on the top. (p12)/(m1) divided by (p22)/(m2) Change divide to multiply sign, flip the second fraction and multiply together to give: (p12x m2)/(m1x p22) Seeing as the momentum's given are the same, they cancel giving (m2)/(m1) Which is answer A. ... phew Edited May 26, 2011 by Sublime Sunshine 2 Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted May 28, 2011 Report Share Posted May 28, 2011 (edited) I have no idea how the answer to this is D : How is the magnitude of the temperature change different Edited May 28, 2011 by timtamboy63 Reply Link to post Share on other sites More sharing options...
Sublime Sunshine Posted May 28, 2011 Report Share Posted May 28, 2011 The change in temperature is different because of the differing specific heat capacities of the two mediums.Consider Q to be the amount of energy which was transferred from the copper block to the steel block:Q = m1c1(delta T1) for the copper blockand similarlyQ = m2c2(delta T2) for the steel blockm1 is the copper block andm2 is the steel block.As Q is the same for these expressions:m1c1(delta T1) = m2c2(delta T2)Seeing as the masses are equal, they cancel and we are left with:c1(delta T1) = c2(delta T2)The specific heat capacities of copper and steel are different, therefore for this equation to be true,(delta T1) cannot be equal to (delta T2)in other words, they must be different. 1 Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted May 28, 2011 Report Share Posted May 28, 2011 But isn't thermal equilibrium when their temperatures are equal? Reply Link to post Share on other sites More sharing options...
Sublime Sunshine Posted May 28, 2011 Report Share Posted May 28, 2011 Yes it is. But the question is asking about the change in temperatures of the two blocks (delta T).This means, (Temperature final) minus (Temperature initial). So while the temperatures end up the same, the amount that they change by through the process of conduction is different. Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted May 29, 2011 Report Share Posted May 29, 2011 Ohh right, got it Thanks! Reply Link to post Share on other sites More sharing options...
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