# Search results

1. ### Two charged spheres hitting each other

ah, of course! I don't know how I didn't see that before, I was thinking I hadn't understood the situation correctly but it was just a typo, many thanks.
2. ### Two charged spheres hitting each other

Thanks for your interest in my question. I had already included my work, the only thing left to do was to show the formula I had derived with the numbers plugged in, which I have now done. The number under the square root is actually positive, according to my calculator.
3. ### Two charged spheres hitting each other

Since the forces involved (gravity and electric force) are conservative we can use conservation of energy. The initial energy is ##E_i= k\frac{q_1q_2}{r_0}-G\frac{m^2}{r_0} ## and the final ##E_f=mv^2+k\frac{q_1q_2}{2r}-G\frac{m^2}{2r} ## so from ##E_i=E_f ## we get...
4. ### Apparent depth of an object underwater

Thanks. I got ##x\approx 18\ m## and an apparent depth of ##d'\approx 0.19\ m##.
5. ### Apparent depth of an object underwater

I would know how to solve this problem if the person had been standing pratically above of the object underwater by using Snell's law and the approximation ##\sin(\theta)\approx\tan(\theta)## fopr ##\theta## small, but in this case I don't see how to find the angles ##\theta_1## and ##\theta_2##...
6. ### Conducting rod in equilibrium due to magnetic force

The problem is that with my ##A=\frac{2Rmg}{hL^2B_0^2}## I get a current...
7. ### Conducting rod in equilibrium due to magnetic force

I get the same result I got in my initial answer: ##i(0)=-\frac{Lh}{R}\frac{dB(0)}{dt}##, ##i(0)LB(0)-mg=0\Rightarrow -\frac{L^2 hB(0)}{R}\frac{dB(0)}{dt}=mg\Rightarrow \frac{dB(0)}{dt}=-\frac{mgR}{L^2 hB(0)}## so ##i(0)=\frac{mg}{LB(0)}##
8. ### Conducting rod in equilibrium due to magnetic force

I don't completely get what you are suggesting but I tried and got: ##iR=-\frac{d\Phi}{dt}=BL\frac{dy}{dt}+Ly\frac{dB}{dt}=Ly\frac{dB}{dt}## and substituting in Newton's second law I would get: iLB-mg=0\Rightarrow -\frac{L^2yB}{R}\frac{dB}{dt}-mg=0\Rightarrow \frac{L^2...
9. ### Conducting rod in equilibrium due to magnetic force

I am having problems understanding point (b) so I would like to know if my reasoning in that part is correct and/or how to think about that part because I don't see how to justify the assumption ##v_y=0\ m/s##. Thanks. I set up the ##xyz## coordinates system in the usual way with ##xy## in the...
10. ### Finding both temperature and the amount of gas added

Thanks; I knew but I was so worried about not being able to understand what was happening in the second part of the problem that I didn't take the time to put the right number of significant figures in my initial calculations. I will obviously correct this when I write up the problem in my notes.
11. ### Finding both temperature and the amount of gas added

The volume of the cylinder is ##V=\pi r^2 h=\frac{7\pi}{250}\ m^3## the number of moles is ##n=\frac{15}{16}\ mol## so from ##PV=nRT## we get ##P=\frac{nRT}{V}=25975.5\ Pa##. Now, for the second question, it should be an isochoric process so ##V_2=V_1## and ##P_2=P_1+0.8P_1=\frac{9}{5}P_1## and...
12. ### Current through ring in solenoid-ring system

The following is my solution to this problem; I would appreciate some feedback, especially on part (b), which I have found the most challenging. Thanks. (a) Using Ampere's Law I get ##B=\mu_0 n i_1## where ##i_1## is the current through the solenoid, and since ##\phi=Li_1##, where ##L## is the...
13. ### Direction of friction for three bodies stacked one on top of the other

I computed (10) again including ##\mu_d## and I got ##a=1.50\ \frac{m}{s^2}## so it appears that indeed ##m_2## is sliding relative to ##m_3##. Thanks again for your help.
14. ### Direction of friction for three bodies stacked one on top of the other

Considering only friction between ##m_3## and the inclined plane intuition tells me that m_2 should start sliding downwards with respect to ##m_3## so that's why the friction force on ##m_2## relative to ##m_3## should point to the left and, by Newton's third law, why the friction force on...
15. ### Direction of friction for three bodies stacked one on top of the other

Isn't that the condition for static friction? I assumed (and the book agrees with me) that the boxes are sliding relative to one another so I used the formula for dynamic friction which, if I am not mistaken, is (in magnitude) constant and equal to ##F_{fr_d}=\mu_d N##.
16. ### Direction of friction for three bodies stacked one on top of the other

The problem is that I thought I had gotten the correct signs for the friction forces between the boxes.
17. ### Direction of friction for three bodies stacked one on top of the other

Thanks, I have corrected that and a couple of other typos.
18. ### Direction of friction for three bodies stacked one on top of the other

I have drawn three free body diagrams, one for each box and then I applied Newton's Second Law after choosing a reference frame rotated clockwise by ##\alpha##, with ##x## pointing south-east and ##y## pointing north-east and I got: ##\begin{cases}m_{1x}: -T+m_1g\sin(\alpha)+F_{fr_{12}}=m_1...
19. ### Moment of inertia of a uniform square plate

Well... doing the integral for the ##n##-th time now I get the correct answer. I don't know why I kept getting it wrong before ... Here it is: ##I=2\int_{x=0}^{x=\frac{L}{\sqrt{2}}}x^2...
20. ### Moment of inertia of a uniform square plate

I placed my Oxy coordinate system at the center of the square, the ##x##-axis pointing rightwards and the ##y##-axis pointing upwards. I divided the square into thin vertical strips, each of height ##h=2(\frac{L}{\sqrt{2}}-x)##, base ##dx## and mass ##dm=\sigma h...

22. ### A spring, disk and pulley system

I think I get it now: when the disk moves leftward for example, it brings up ##R\theta## of the rope and its axis of radius ##r## "eats" ##r\theta## of the rope (because the rope is winding around it) so the combined effect on ##m## is that it must go upward of an amount ##s=(R+r)\theta## and...
23. ### A spring, disk and pulley system

Thinking about it, since the rope is attached to axis of radius ##r## orthogonal to the disk, if the disk moves a distance ##x## then, since it moves without slipping it must be ##x=R\theta## where ##\theta## is the angle the disk (and also the axis) turn and so the length of the rope changes...
24. ### A spring, disk and pulley system

(a) By setting up a coordinate system with the x-axis pointing to the right and the y-axis pointing downward we have ##\begin{cases}-kx_{eq}+T_1+F_{s}=0\\ -RF_{s}+rT_1=0\\ r_p (T_2-T_1)=0\\ -T_2+mg=0\end{cases}\Rightarrow x_{eq}=\frac{mg}{k}\left(1+\frac{r}{R}\right)## which coincides with the...
25. ### Image position and magnification for underwater spherical lens

https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/02%3A_Geometric_Optics_and_Image_Formation/2.04%3A_Images_Formed_by_Refraction#Equation+20:+Refraction+at+a+convex+surface
26. ### Maximum charge on the plates of a capacitor

##\phi=\int_{S}\vec{B}\cdot d\vec{S}## and since ##\vec{B}## is pointing inside the page and the area is oriented with the normal pointing away from the page this becomes ##\int_{S}(-B)dS=-B\int_{S}dS=-Ba^2## so ##\mathcal{E}=-\frac{d}{dt}\phi=-\frac{d}{dt}(-Ba^2)=a^2\frac{dB}{dt}.##
27. ### Maximum charge on the plates of a capacitor

What I have done: The electromotive force due to Faraday's Law is: ##\mathcal{E}=-\frac{d\phi(\vec{B})}{dt}=\frac{d}{dt}(Ba^2)=a^2\frac{dB}{dt}=-10^{-4}V.## In the circuit, going around the loop in a clockwise fashion: ##\oint_{\Gamma}\vec{E}\cdot d\vec{l}=-\frac{d\phi(\vec{B})}{dt}\Rightarrow...
28. ### Image position and magnification for underwater spherical lens

What is it that you think I should change in my solution? I too was confused by the discrepancy between the diagram and the text (the glass is blue and the water is gray...) but in the end I decided to follow the diagram, hence my solution.
29. ### Image position and magnification for underwater spherical lens

Yes, I think it does. The problem confuses me a bit because I don't know if I can assume that the glass lens will behave like a mirror or not. If it does, looking at a similar configuration here (https://opentextbc.ca/universityphysicsv3openstax/chapter/spherical-mirrors/) I would say yes.
30. ### Image position and magnification for underwater spherical lens

Using the data given and recalling that in this configuration ##R<0## I get: ##\frac{1.33}{0.5}+\frac{1.5}{q}=\frac{1.5-1.33}{-0.2}\Rightarrow q\approx -0.427 m=-42.7 cm## so the image is virtual and is ##42.7\ cm## to the left of vertex ##V##. The magnification is ##M=\frac{n_1 q}{n_2...
31. ### Change of entropy in the Universe in a thermodynamic cycle

##\eta=\frac{100,000}{1800000}=\frac{1}{18}<\frac{19}{20}=1-\frac{T_{cold}}{T_{hot}}=\eta_{Carnot}.##
32. ### Change of entropy in the Universe in a thermodynamic cycle

But now, the efficiency should be ##\eta=\frac{W_{done}}{Q_{absorbed}}=\frac{300,000}{1800000}=\frac{1}{6}<\frac{19}{20}=1-\frac{T_{cold}}{T_{hot}}=\eta_{Carnot}## so it is still not a reversible cycle, right?
33. ### Change of entropy in the Universe in a thermodynamic cycle

Thank you very much for your detailed answer. The work is to be intended as the work done ON the gas, so ##W_{B\to C}## should have been ##W_{B\to C}=-300,000\ Joule## and now the total work done by the gas during a cycle is equal to the heat added to the gas during a cycle. Regarding the...
34. ### Change of entropy in the Universe in a thermodynamic cycle

(a) We first find that: ##T_A=\frac{P_A V_A}{nR}=\frac{1\cdot 10^5 \cdot 4}{40\cdot 8.314}K\approx 1202.7904 K##, ##\frac{T_B}{T_A}=\frac{\frac{P_B V_B}{nR}}{\frac{P_A V_A}{nR}}=\frac{P_B V_B}{P_A V_A}=\frac{P_A \frac{V_A}{2}}{P_A V_A}=\frac{1}{2}##, ##\frac{T_C}{T_B}=\frac{P_C...
35. ### Number of bright fringes given by diffraction grating on a screen

Thanks, I forgot to divide by ##L##, it should be correct now.
36. ### Number of bright fringes given by diffraction grating on a screen

Ah, I see, so if I understand correctly since ##\sin(\theta_m)=m\frac{\lambda}{d}## for the infinite screen I have to impose that ##\sin(\theta_m)\leq 1\Leftrightarrow m\frac{\lambda}{d}\leq 1\Leftrightarrow m\leq \frac{d}{\lambda}## so I can get at most ##\frac{d}{\lambda}## maxima hence...
37. ### Number of bright fringes given by diffraction grating on a screen

(1) In the book I am using the separation of bright fringes is indicated as being ##\Delta y=\frac{\lambda}{d}##, where ##d## is the separation of the slits so on a screen of width ##W## I would see ##\frac{W}{\frac{\lambda}{\frac{1}{N}}}## bright fringes. I don't see why the text of the...
38. ### Force to apply to a loop moving away from a current-carrying wire

I see, thank you very much.
39. ### Force to apply to a loop moving away from a current-carrying wire

I see, thanks. So, if I understand correctly, if I treated ##h## as a constant (representing the initial loop-wire distance at ##t=0##), ##F=\frac{\mu_0^2i^2l^4Nv}{(2\pi)^2Rh(h+l)(h+vt)(h+l+vt)}## would be a correct formulation for the force to be applied at time ##t##, right?
40. ### Force to apply to a loop moving away from a current-carrying wire

What I have done: (1) ##\Phi(\vec{B})=\int_{S}\vec{B}\cdot d\vec{S}=-\frac{N\mu_0 il}{2\pi}\int_{s=h}^{s=h+l}\frac{ds}{s}=-\frac{\mu_0iNl}{2\pi}\ln(\frac{h+l}{h})## so ##\mathcal{E}=-\frac{d\phi(\vec{B})}{dt}=-\frac{\mu_0iNl^2v}{2\pi h(h+l)}## so...
41. ### Time for current in solenoid to be 10% less than the steady state current

Thank you very much, no wonder I couldn't see what I was doing wrong.
42. ### Time for current in solenoid to be 10% less than the steady state current

The mistake was purely mine, thanks. Nonetheless the book claims the answer should be ##0.16s##.
43. ### Time for current in solenoid to be 10% less than the steady state current

there was a typo in my answer (now corrected): the ##t## was part of the exponent and so the formula I had obtained should be correct.
44. ### Time for current in solenoid to be 10% less than the steady state current

I set up the equation ##V-iR-L\frac{di}{dt}=0##, with ##i(0)## and by solving it I got ##i(t)=\frac{V}{R}(1-e^{-\frac{R}{L}t})##. Then, since the steady state current is ##i_s=\frac{V}{R}## I imposed the condition ##i(t_1)=\frac{9}{10}\frac{V}{R}\Leftrightarrow...
45. ### Finding equilibrium temperature when there are phase changes

If there weren't phase changes occurring I know that the temperature equilibrium would be ##T_e=\frac{m_{ice}c_{ice}T_{ice}+m_{w}c_{w}T_{w}}{m_{ice}c_{ice}+m_{w}c_{w}}##. Now, by repeating the reasoning to get the above formula (##\sum \Delta Q=0##) and adding the phase changes of the water...
46. ### Find the time for the current to fall to 35mA in this simple RL circuit

I understand now, thanks. It should have been ##i(t)=0.06e^{-(R/L)t}## since, from time ##t=0## onwards, there isn't a battery connected to the "new" circuit anymore. By setting ##i(t^*)=35\cdot 10^{-3} A## I now get the correct value, thanks.
47. ### Find the time for the current to fall to 35mA in this simple RL circuit

What I have done: (1) ##I(0)=\frac{V}{R}=\frac{1.5}{25}A=0.06 A.## (2) By setting ##I(t*)=0.06(1-e^{-(35/0.4)t*})=35 mA## we get ##t*\approx 0.01 s## What I have done seems correct to me, but the result for part (2) should be different. I would be grateful if someone could point out to me...
48. ### Flux of constant magnetic field through lateral surface of cylinder

Ah, thinking more about it, could I say that since the ##\vec{B}##-field is constant across space and since the cylinder is symmetric I would have that the flux through the lateral surface of the cylinder is ##0##, because when "entering" the cylinder it has a plus sign and when "exiting"...
49. ### Flux of constant magnetic field through lateral surface of cylinder

That the flux through one is the opposite of the flux through the other in the case that I have described where the axis of the cylinder lies on the line spanned by the vector ##(3,2,1)## but can I say anything more without knowing something more about the orientation of the cylinder in space?
50. ### Flux of constant magnetic field through lateral surface of cylinder

If the question had been asking about the flux through the whole surface of the cylinder I would have said that the flux is 0, but since it is asking only about the lateral surfaces I am wondering how one could calculate such a flux not knowing how the cylinder is oriented in space. One could...