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Update, got the results yesterday and I got half the points for this problem! :-) Thanks for helping!

Dear all, I officially passed all of my exams yesterday. It seemed very unlikely but now I get to study physics and hopefully pursue my childhood dreams. I want to really thank you. I quit my job and was on a budget so that I couldn't afford tutoring, but I don't think I would have been able to...

Hi Kuruman, thank you for answering me again. That makes sense. The loop is originally falling vertically down perpendicular to the magnetic field. It's why when I thought about it I imagined the dipole being created inside the loop, directly opposite the direction of the magnetic field which...

Hi all, I'm sorry I just thought about this problem a little more and have another question. Though it is a square, can I not see this wire as a loop and thus, wouldn't it also create a magnetic dipole that could end up feeling a torque? Because by lenz's law the current induced should be in...

Hi mjc123 and drclaude. Thank you for explaining. I have come across the equipartition theorem in halliday's textbook (which I got after this problem because I haven't come across the various C_v before). I think I have a better idea now, but I'll have to ponder on it a bit more to really...

Hi, thanks drclaude. I'm sorry if I'm misunderstanding something here. I was just wondering since the kinetic energy is ## \frac {1}{2}m(v_rms)^2 = \frac {3}{2}kT ## it seemed to me like temperature is only dependent on kinetic energy. And I was wondering why there isn't accounted for...

Hi mjc123, sorry to come back to this thread again after such a long time. thanks to you and chestermiller I think I have a real good grasp and can use these formulas well (I did a lot of problems I found in two textbooks). I still have a concept question though and it might be a silly...

Hello TSny. Thank you very much. I lost quite a bit of time there. (trying to find a formula to make the masses cancel) And I thought about it for the past two days without figuring it out. I thought about it and with the density I probably could have figured it out. (Volume being 4a times A)...

Hi all, so I had this problem and on the exam and I got a solution but I had an mass-term in there which wasn't given. I used Farraday's Law of Induction to get the Voltage induced. Then I used ##rho* \frac{A}{4a} ## for the resistance and divided the Voltage by that to get the current. I then...

Omg! You're right doggydan42, I evaluated e^0 as 0 instead of 1. I made this mistake each time I solved it. It's so frustrating, these simple things seem to still happen to me so often. (This morning I was working on another problem where I kept 2*1 = 3 and couldn't figure out what I was doing...

Hi doggydan42, could it be that the differential equation should be : ## V_0 = -\frac{dV}{dt}RC + V ## ?

Hi @doggydan42 thank you. I tried to make a clean version of what happened. I didn't upload the work previously because once i get confused it becomes a mess that only I can read. First is the integrating factor. I was able to solve the latter for the right solution, though if I use -dV/dt for...

I think I might be making an integration mistake. Is ## V_0 = CR*\frac{dV}{dt} +V ## the correct Differential equation? edit: I am definitely making integration mistakes. Sorry! I thought I had to use the integrating factor. And then used u sub. But with some algebra now, I could do without and...

Hi vela, thank you very much. It seems like your earlier comment put me on the right track but I messed up somehow. I tried replacing I in IR through and the Resistance canceled and I ended up with the V_0 = V expression. I will try to think about it some more and see if I can't come up with...

Hi @vela thank you very much. the current was more intuitive, I knew it's dq/dt so it just came. I did some algebra (Q to the dq/dt side) and then I integrated and then raise to the power of e to get Q out of the natural log to get the Result. dq/dt is changing due to the amount of charge on...

Hi Vela, thank you very much for helping. I thought about your comment and tried a few ways earlier. Didn't quite seem to find a solution. I came back after reviewing basics but I'm still stuck. The current decreases as the Voltage (Electric PE) increases in the capacitor. So I thought that...

Hi all, thank you very much for your help. At first I felt confident after@doggydan42's suggestion and tried last night, but I had problem identifying my ## \frac {dV}{dt} ## part. I did a dimensional analysis as suggested by @DaveE of my solution and got Coloumbs * Ohms over seconds which...

Hi doggydan42, thank you very much. I thought of this, but for some reason I thought it's not ok to just use V in there. But it makes sense. Let me try again, now I'm also curious if I'll get a similar derivative to the one I got doing my long and tedious way! thanks so much!

Hi all. I have another exam question that I am not so sure about. I've solved similar problems in textbooks but I have a feeling once again that the correct way to solve this problem is much simpler and eluding me. Especially because my answer to a) is already the solution to c) and d) (I did...

ω Hi Orodruin. Thank you, this is an identity I didn't know, or remember! I'm glad that this means I must have just made an algebra or calculator mistake. Thank you very much! Edit: I've just drawn it out and it's actually so obvious that they are the same! thanks again!

Hi all, I found this problem in a new textbook I'm working through. And my energy conservation equation was ## mg\frac {h}{2} = \frac {1}{2} I ω^2 + mg \frac {h}{2}*sin(55) ## My solution was wrong and after checking why I found that they used cos(35) as the angle. The rest was the same. I'm a...

Hi @kuruman I wanted to let you know that it sank in yesterday and I feel I understand the benefit and why to chose the coordinate axes like you suggested. I've been doing lot's of problems and I tried to follow your suggestions but it hasn't quite made click, even though I could always get the...

Thank you! It took a while to wrap my head around it but I can completely follow now what you did. Thank you for your patience. I was especially slow today (hopefully it's just the heat!) and yet I learned a new relation that I think will come in handy!

Thank you, this sounds very intriguing. I wasn't able to find any textbook that explained this way of calculating C_p yet. But I'm going to search further. Thanks again.

Thank you. I saw what you did in the first part and I could follow. I have to try to reason a bit on how we arrived at the density, I'm don't follow this quite yet.

Thank you chestermiller. May I ask if you calculated the density and if so how? All I knew how to do was to look it up in a textbook table.

Thank you chestermiller. There might be a misunderstanding. I did use the correct formula and R. And I've gotten the same result as you. I mentioned this above, but I probably wasn't clear enough, I'm sorry. I tried to verify that result ## \Delta T= 87.2C ## by using ## Q=m*c* \Delta T## and...

Thank you chestermiller for answering. I'm sorry I don't quite understand? Do you mean there is a way to equate these two and arrive at a solution? I didn't use these in the same equation. Rather in two ways, in which I thought I should arrive at the same solution. The Temperature Difference...

Hi I solved it with the Molar Heat Capacity thanks to mjc123's help and got a difference of 87.2 Degrees. I've been trying yesterday and today morning to verify my results with ##m*c*(T_1-T_2)## and ##3/2*n*R*(T_1-T_2)## For the former, I used the density of ## 1.293 \frac {kg} {m^3} ## For R...

Hi mjc123, So I worked out the problem and I got around 107 Degrees C. Which seems about right (and I get almost the same by looking up the average density of the air and multiplying by Volume to use mc(T_1-T_2). Just plugging in the numbers my first intuition is definitely wrong and very...

Hi mjc123. Thank you, that is very helpful to know. I don't remember in which textbook I got it (I think the molar heat capacity way wasn't in many) from but I'll search and find it again to commit to memory.

Hi mjc123 Thank you for answering. My first intuition was actually to use the Ideal Gas Law and then go with the heat capacity. The number of mols are 0.394... by my calculation. I was able to look up the heat capacity. It's 1.4 Edit! (1.4*3/2*R)? But I wouldn't have been able to know it by...

Hi, so I found this on another old "AP" High School Finals Exam. I think I may be super lost. Because the only way that I can think about is KE = 3/2kT. And then that the difference of the Kinetic Energy of the air Particles is the Q supplied by the heater inside the air dryer. So ## \frac...

Hi Kuruman, Thank you. Seeing that the term cancels has helped. I will try to rework the problem to let it sink in and come back.

Hi Kuruman, thank you. I have trouble. I'm sorry. I drew it just now, but... I've worked with your reasoning, up until ##\Delta U=\Delta U'_{spring}## Once Mass and spring move from said zero (equilibrium), why do we not account for the change in potential gravitational energy? Please believe...

Thank you!

Thank you, Kuruman. This is very helpful, and I think I did have this understanding but I may misunderstand the big picture. I understand why ##ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2##. But, I'm probably asking a really dumb question that is already explained with your reasoning...

I think it is not, so I adjusted for that. Otherwise, A would be 0.50 m no? Instead of 0.206 and x of mgx would be -0.50 instead of -0.206 I'm sorry if I'm somewhat slow to catch on.

Hi Kuruman! Thank you for helping me on this post as well. Originally I only had w as L when I first worked on the problem. Because only this section of the wire experiences a net force. But since the current is induced through all the wire I thought I should use the length that is exposed to...

Hi Kuruman, thank you. I reread your post. I thought that I had used the equilibrium position as zero for the gravitational potential energy?

Hi, second problem in one evening, I'm sorry! But I'm also not quite sure if I did this one right. I had thought I need lenz's law but there is no current before entering the field so I just use the induced Voltage? My approach: ## V = \frac {B*A}{t} ## ## IR = \frac {B*A}{t} ## and ## A = v*t...

Dear all, I am back with another Spring problem. I have tried to use the insights I gained from your help last time: https://www.physicsforums.com/threads/find-v-x-of-a-mass-suspended-from-a-spring.972942/page-3#post-6190934 I figured I start with calculating the new equilibrium by ## x...

Oh, thanks. Of course this makes absolute sense. Thank you !

Hi, I have a conceptual question. I was doing some problems on pendulums and found something that confused me. I attached a drawing. I used to always solve these problems by using some trigonometry and trying to find the Tension. i.e. ## m*g = F (of the Tension) * cos(theta) ## so ## \frac...

Hi, Thank you! I love the openstax books, they helped me a lot, especially in math and in particular their wealth of problems. I was really silly. I have worked on the Openstax College Physics book and found that they don't have solutions, and the students' solutions were only minimal, unlike...

@Doc Al Thank you. The spring did it for me!

Hello @Doc Al , Thank you very much for answering! The correct approach has been the most obvious. I'm trying to do old problems in several ways now to practice. I'm sorry I don't understand yet. Why would the charge swing past the equilibrium point? Maybe I'm picturing it wrong. I imagined...

Hi, so I was able to solve this problem by just equating the forces (Tension, mg, and EQ). But I thought I could also solve this problem with Conservation of Energy. However, I calculated it several times, and I never get the right answer this way. Doesn't the Electric Field do the work to put...

Hi all, can you recommend a textbook or other resource with comprehensive (meaning spanning multiple subjects) and challenging physics problems and also the solutions? I have already worked through Giancoli (no-calculus), Ohanian (calc based) and the two dummies workbooks. I have been working...

Hi, Thanks. I used the formula from that link and got a different speed. Around 19, whereas my solution gets about 27 meters per second. I probably made a mistake in plugging in the numbers. Thank you so much for checking!