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.
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.
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...
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##...
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)}##
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...
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...
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.
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...
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...
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.
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...
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##.
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...
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...
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...
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...
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...
(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...
##\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}.##
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...
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.
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.
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...
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?
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...
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...
(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...
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?
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...
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...
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...
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.
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...
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"...
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?
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...