Physics Problems Q&A - Recent questions without answers
http://physics.qandaexchange.com/?qa=unanswered
Powered by Question2Answertime required to pass electricity through electroplating bath
http://physics.qandaexchange.com/?qa=3447/time-required-pass-electricity-through-electroplating-bath
<p>The time required to pass 3600 Coulomb of electricity through an electroplating bath using a current of 200 mA is <strong><strong>__</strong></strong>_ hours</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3447/time-required-pass-electricity-through-electroplating-bathWed, 17 Apr 2019 03:09:40 +0000Container filled with fluid move with a acceleration a
http://physics.qandaexchange.com/?qa=3446/container-filled-with-fluid-move-with-a-acceleration-a
<p><img src="https://i.imgur.com/2lDHr3Z.jpg" alt=""></p>
<p>My try : </p>
<p><img src="https://i.imgur.com/5rhhb1t.jpg" alt=""></p>
<p>My option B is not getting matched.<br>
Answer is given as B, C.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3446/container-filled-with-fluid-move-with-a-acceleration-aTue, 16 Apr 2019 15:23:24 +0000Given the vector current density, determine the total current flowing outward through a circular band
http://physics.qandaexchange.com/?qa=3431/current-density-determine-current-flowing-outward-circular
<p>Given the vector current density, determine the total current flowing outward through a circular band.:</p>
<p><a rel="nofollow" href="https://i.stack.imgur.com/klQAo.png"><img src="https://i.stack.imgur.com/klQAo.png" alt="enter image description here"></a></p>
<p>The answer should be 518A. It comes something around 3255 A. Where is the mistake?</p>
<p><a rel="nofollow" href="https://i.stack.imgur.com/g1T1l.jpg"><img src="https://i.stack.imgur.com/g1T1l.jpg" alt="enter image description here"></a></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3431/current-density-determine-current-flowing-outward-circularFri, 29 Mar 2019 11:20:51 +0000Nucleon-Nucleon Scattering
http://physics.qandaexchange.com/?qa=3430/nucleon-nucleon-scattering
<p><img src="https://i.imgur.com/hZhT97F.png[/img]" alt=""><br>
<img src="https://i.imgur.com/rklrj0w.jpg[/img]" alt=""></p>
<p>I computed the orbital angular momentum classically, and then used quantum mechanics. But I am not convinced of what I got, because $L \alpha \sqrt{E}$ is a proportional equation that does not justify that when E < 20 MeV you get L ~ 0 (I do not see, numerically speaking, the difference between either plugging 20 MeV or 19MeV).</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3430/nucleon-nucleon-scatteringWed, 27 Mar 2019 12:38:33 +0000Where do I make the approximations in this proton and electron collision problem?
http://physics.qandaexchange.com/?qa=3407/where-make-approximations-proton-electron-collision-problem
<blockquote><p>A proton moving with a velocity $\beta c$ collides with a stationary electron of mass $m$ and knocks it off at an angle $\theta$ with the incident direction. Show that the energy imparted to the electron is approximately $$T \approx \frac{2\beta^2 \cos^2\theta}{1-\beta^2\cos^2\theta}mc^2$$</p>
</blockquote>
<p><em>Link to question :</em><br>
<a rel="nofollow" href="https://www.chegg.com/homework-help/questions-and-answers/proton-moving-velocity-collides-stationary-electron-mass-m-knocks-angle-incident-direction-q35643739?trackid=VRP2b2Jt">https://www.chegg.com/homework-help/questions-and-answers/proton-moving-velocity-collides-stationary-electron-mass-m-knocks-angle-incident-direction-q35643739?trackid=VRP2b2Jt</a></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3407/where-make-approximations-proton-electron-collision-problemSun, 17 Mar 2019 04:34:00 +0000Griffiths 5.4 Force on a square loop
http://physics.qandaexchange.com/?qa=3404/griffiths-5-4-force-on-a-square-loop
<p><img src="https://i.imgur.com/HckbPJf.png[/img]" alt=""></p>
<p><img src="https://i.imgur.com/9SEg3Li.png[/img]" alt=""></p>
<p>Before doing any calculation, I see the net force being zero;</p>
<p>$$ F_{AB} = -F_{CD}$$</p>
<p>$$ F_{BC} = -F_{DA}$$ </p>
<p>But apparently this is not the case. The provided solution is:</p>
<p><img src="https://i.imgur.com/75ijztI.png[/img]" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3404/griffiths-5-4-force-on-a-square-loopSat, 09 Mar 2019 11:53:06 +0000question about 4-velocity
http://physics.qandaexchange.com/?qa=3396/question-about-4-velocity
<p>A particle is moving in the x, y plane at a speed of v = 0.80 and it is travelling an angle of 60 degrees<br>
above the x-axis.</p>
<p>(a) Rotate the spatial coordinates so that v lies along the x-axis and then construct the components of the 4-velocity for the particle.</p>
<p>(b) Construct the remaining orthonormal basis vectors for the particle in the rotated frame, then rotate the spatial coordinates back to find the basis vectors in the original frame.</p>
<p>(c) Perform the same steps above but this time begin by rotating the spatial coordinates so that <br>
v lies along the y-axis. Do you get the same result for the basis vectors after rotating back to the original frame? Should you? Draw a picture with ~v, the original (x, y) frame, and the two rotated frames to explain what is happening.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3396/question-about-4-velocityWed, 27 Feb 2019 22:32:29 +0000Frequency modes of the rectangular shell
http://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shell
<p>This is task i received from my professor:</p>
<blockquote><p>The shapes of three natural modes having the frequencies $\omega_1, \omega_2, \omega_3$ of the rectangular shell are presented in the figure. The exciting pressure $p(t)$ applied uniformly all over the one side of the shell has the form $p(t) = Pe^{jωt}$. <br>
Make a sketch of the normal displacement of the gravity point of the shell against frequency, if the excitation frequency varies within bounds $0.5\omega_1< \omega <2\omega_1$ and static displacement of that point equals to $u_0$.</p>
</blockquote>
<p>Links to photo of frequency nodes (sorry for low quality)-><a rel="nofollow" href="https://ibb.co/b1nh3j9">1</a> .</p>
<p>Can somebody help me and tell me how i should get started?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shellMon, 14 Jan 2019 21:42:44 +0000Wave optics- number of minimas.
http://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimas
<p><img src="https://cdn.pbrd.co/images/HUWujTQ.jpg" alt=""></p>
<p>Attempt: </p>
<p>Let m be the integer associated with $420 nm$<br>
and n be the integer associated with $540 nm$</p>
<p>$d \sin \theta = k \lambda $ $k\in Z$<br>
Clearly, </p>
<p>$m_{max} = 180$ </p>
<p>$n_{max} = 140 $</p>
<p>$(2m+1) \lambda_1 = (2n+1)\lambda_2$ (condition for dark fringes to overlap)</p>
<p>$\implies \dfrac{2m+1}{2n+1} = \dfrac 97$</p>
<p>$\implies m = \dfrac{1+7n +2n }{7}$</p>
<p>Hence, we obtain, for m to be an integer: $2n+1 = 7k$</p>
<p>$\implies n = \dfrac{7k-1}{2}$ where $k \in Z$</p>
<p>Now note that k must be odd since odd-1 = even </p>
<p>Thus, using $n \le 140$, $k_{max} = 39$</p>
<p>Now, we have to consider only odd values of k which are $1,3,...39$ = 20 numbers </p>
<p>Thus, we have 20 minimas on the upper side and 20 on the lower,<br>
Total minimas = $20+20 = 40$</p>
<p>But answer is $D$</p>
<p><strong>Question 1:</strong> What is wrong with my method? </p>
<p><strong>Question 2:</strong> Considering that this is a JEE Mains problem, what is the fastest 2 minute way to do it? </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimasFri, 04 Jan 2019 22:01:11 +0000Liquid in a capacitor as dielectric (IE Irodov 3.144)
http://physics.qandaexchange.com/?qa=3225/liquid-in-a-capacitor-as-dielectric-ie-irodov-3-144
<blockquote><p>A parallel plate capacitor is located horizontally so that one of its plates is submerged into the liquid while the other is over its surface. The permittivity of the liquid is equal to $\epsilon$, its density is equal to $\rho$. To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface charge density $\sigma$?</p>
</blockquote>
<p>This is a type of controversial problem, as it contains a different answer by the different author, two different answers provided for this are </p>
<p> $$1) \ h=\dfrac{(\epsilon-1)\sigma^2}{2\epsilon_o\epsilon\rho g}$$ $$2) \ h=\dfrac{(\epsilon^2-1)\sigma^2}{2\epsilon_o\epsilon^2\rho g}$$</p>
<p><a rel="nofollow" href="https://ibb.co/fYmek0">Solution for 1</a> and <a rel="nofollow" href="https://youtu.be/oyfo-TJEG28?t=31">solution for 2</a></p>
<p>Please help!</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3225/liquid-in-a-capacitor-as-dielectric-ie-irodov-3-144Tue, 30 Oct 2018 01:37:42 +0000induced charge density at boundary surface
http://physics.qandaexchange.com/?qa=3216/induced-charge-density-at-boundary-surface
<blockquote><p>The space between the plates of a parallel plates capacitor is filled consecutively with two dielectrics layers $1$ and $2$ having the thickness $d_1$ and $d_2$ respectively and permittivities $\epsilon_1$ and $\epsilon_2$. The area of each plate is equal to $S$ find:<br>
1. The capacitance of the capacitor <br>
2. The density $\sigma'$ of the bound charges on the boundary plane if the voltage across the capacitor equals $V$ and the electric field is directed from layer $1$ to layer $2$.</p>
</blockquote>
<p>I managed to find $C_{eq}$ easily, for the second part I don't understand that word " <em>boundary plane</em>", I only know $\sigma'=\sigma\bigg(1-\dfrac{1}{\epsilon}\bigg)$ holds when there would have been just one plate, how to work on two such consecutive plates. Please help.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3216/induced-charge-density-at-boundary-surfaceSun, 28 Oct 2018 05:02:22 +0000Deducing the mass of the Sirius system - Feynman exercises 3.17
http://physics.qandaexchange.com/?qa=3185/deducing-the-mass-of-the-sirius-system-feynman-exercises-17
<p><img src="https://cdn.pbrd.co/images/HJlk9yQ.jpg" alt=""></p>
<p>I do not even know how to try to deduce the mass M of the Sirius system in terms of that of the sun, as I do not not how to interpret the given data. </p>
<p>Could you give me a hint? </p>
<p><strong>EDITED AT THIS POINT</strong></p>
<p><strong>1) Estimating the distance to the Sirius binary system.</strong></p>
<p>I calculated the hypotenuse of the triangle $S E_1 SS$ (Sun - position of the Earth located at the top of a vertical circular orbit - Sirius System):</p>
<p>$$ \theta = 0.378^{\circ} arc \times \frac{1}{3600} \times \frac{2 \pi}{360} = 1.83 \times 10^{-6}arc sec $$</p>
<p>$$D = \frac{D'}{sin(\theta)} = 8.16\times 10^{16} m$$</p>
<p>Where:</p>
<p>D = the distance from the Earth to the Sirius binary system.</p>
<p>D' = Distance from the Sirius system to the Earth ($15 \times 10^{10} m$).</p>
<p>Here's an illustration of what I have done:</p>
<p><img src="https://cdn.pbrd.co/images/HJtrflM.png" alt=""></p>
<p><strong>2) Measuring the semi-major axis of the ellipse and estimating the period of the orbit from the figure.</strong></p>
<p>To compute the semi-major axis I have assumed the major axis is under the scale [0'',12'']. </p>
<p>As Sammy Gerbil said :'To measure the semi-major axis from the figure, magnify the diagram to fill your screen, place the edge of a sheet of paper along the diagonal line marked on the ellipse, mark off the ends of this line on the paper, then transfer the paper edge to the scale and read off the length in arc seconds. I get 13.8" for the major axis length, so the semi-major axis is a=6.9". The calculation is then:'</p>
<p>$$a=8.16\times 10^{16}m\times \frac{6.9}{3600}\times \frac{2\pi}{360} = 2.73 \times 10^{12}m$$</p>
<p>Now let's estimate the period of the orbit.</p>
<p>Is this the right thought? </p>
<p>$$\frac{dA}{dt} = SM \times SM \times \frac{d \theta}{2dt}$$</p>
<p>It will not be accurate as 'the two sides of most of the triangles are different'.</p>
<p>$A_1$:</p>
<p><img src="https://cdn.pbrd.co/images/HK5k6ab.jpg" alt=""></p>
<p>On process...</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3185/deducing-the-mass-of-the-sirius-system-feynman-exercises-17Sat, 20 Oct 2018 15:50:53 +0000Find the angle between the $x$-axis and a vector
http://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vector
<blockquote><p>The $x$ component of vector $A$ is 25.0 m and the $y$ component is 40.0 m. <br>
(a) What is the magnitude of $A$? <br>
(b) What is the angle between the direction of $A$ and the positive direction of x?</p>
</blockquote>
<p>For (b) I tried using the formula $\tan \theta = \frac{a_y}{a_x} = \frac{40}{-25} = -1.6$, thus $\arctan(-1.6)=58$ degrees which does not match the answer key: $122$ degrees.</p>
<p>Any help is appreciated.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vectorSat, 20 Oct 2018 11:59:19 +0000Normal reaction for particle at rest on a sphere
http://physics.qandaexchange.com/?qa=3171/normal-reaction-for-particle-at-rest-on-a-sphere
<p><img src="https://cdn.pbrd.co/images/HIVCKHf.png" alt=""></p>
<p>The answer is C. </p>
<p>It is not difficult to understand why force $F$ decreases as the small metal ball is raised to the top of the sphere. The normal reaction $R$ from the hemisphere bears a greater share of the weight of the metal ball. </p>
<p>But why is the normal reaction $R$ the same for all positions of the ball on the sphere? Is there an intuitive explanation?</p>
<p>See discussion in <a rel="nofollow" href="https://chat.stackexchange.com/transcript/message/47213469#47213469">Problem Solving Strategies chatroom</a> on Physics Stack Exchange.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3171/normal-reaction-for-particle-at-rest-on-a-sphereWed, 17 Oct 2018 22:19:45 +0000Angle of pendulum in accelerated incline
http://physics.qandaexchange.com/?qa=3107/angle-of-pendulum-in-accelerated-incline
<p>Consider the following system consisting of a box sliding down a plane. The coefficient of friction between the plane and the box is $\mu$. A pendulum is attached to the top of the box as shown.</p>
<p><img src="http://i65.tinypic.com/1z37azr.png" alt=""></p>
<p>The acceleration of the box+pendulum is $a=g(\sin\theta-\mu\cos\theta)$ I believe.<br>
In a non-inertial frame attached to the box, the free-body diagram for the pendulum is</p>
<p><img src="http://i63.tinypic.com/2eyyx3s.png" alt=""></p>
<p>My goal is to find the angle $\phi$ from the equilibrium of these 3 forces. I have to pick x and y axes to decompose these forces. If I pick the x axis along the fictitious force and the y perpendicular to it I get (I think) </p>
<p>$$m_P a+T\sin\phi=m_P g\sin\theta$$ </p>
<p>and </p>
<p>$$T\cos\phi=m_{P}g\cos\theta$$</p>
<p> which I can then solve for $\phi$ :</p>
<p>$$\tan\phi=\frac{g\sin\theta-a}{g\cos\theta}=\mu$$</p>
<p>Is this right?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3107/angle-of-pendulum-in-accelerated-inclineFri, 28 Sep 2018 04:00:23 +0000$\Psi$ in function of 'position'
http://physics.qandaexchange.com/?qa=3028/%24-psi%24-in-function-of-position
<p>The general exercise:<br>
<img src="https://cdn.pbrd.co/images/HApljRF.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HAQZH0G.jpg" alt=""></p>
<p>Let's focus on c). </p>
<p>The value of $\sigma$:</p>
<p>$$\sigma =\sqrt{ < x^2 > -< x >^2 }= \frac{1}{\lambda \sqrt{2}}$$</p>
<p>We are asked to plug in $\sigma$:</p>
<p>$$| \Psi (\sigma ) | = \lambda e^{\frac{-2}{\sqrt{2}}}$$</p>
<p>Where $\lambda$ is a positive constant.</p>
<p>I calculated < x > and its value is 0. Check this link to obtain further details:</p>
<p><a rel="nofollow" href="https://math.stackexchange.com/questions/2893056/misleading-expected-value">https://math.stackexchange.com/questions/2893056/misleading-expected-value</a></p>
<p><img src="https://cdn.pbrd.co/images/HBKGTII.png" alt=""></p>
<p><strong>About the probability</strong></p>
<p>I have difficulties solving this probability as this time I got:</p>
<p>$P_i = \int_{-\sigma}^{\sigma} \lambda e^{-2 \lambda |x|} dx =$ </p>
<p>$= \lambda \int_{-\sigma}^{0} e^{-2 \lambda |x|} dx$ </p>
<p>$+ \lambda \int_{0}^{+\sigma} e^{-2 \lambda |x|} dx = 0$</p>
<p>Where did I get wrong? I have been trying to get this probability but none outcome seems to make sense.</p>
<p>$$P_o = 1 - P_i $$</p>
<p>Where $P_i$ and $P_o$ mean inside and outside respectively</p>
<p>Thank you</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3028/%24-psi%24-in-function-of-positionSat, 25 Aug 2018 21:27:13 +0000Both molar specific heat of an ideal monatomic gas at constant volume and pressure
http://physics.qandaexchange.com/?qa=2984/both-molar-specific-ideal-monatomic-constant-volume-pressure
<blockquote><p>2 mols of oxygen are heated up from a temperature of 20 C and a pressure 1 atm to a temperature of 100 C. Compute:<br>
a) Having constant volume, how much heat must be provided to the gas throughout the process?<br>
b) Having constant pressure, how much work is done throughout the process?<br>
c) Compute in a) and b) cases the variation in the internal energy.</p>
</blockquote>
<p>My try:</p>
<p>The molar specific heat of an ideal gas at constant volume is:</p>
<p>$$C_v = \frac{3}{2}R$$</p>
<p>The molar specific heat of an ideal gas at constant pressure is:</p>
<p>$$C_p = \frac{5}{2}R$$</p>
<p>Then at a) it is just about using $Q = nC\Delta T$. My issue here is that the given outcome at a) uses n = 1:</p>
<p>$$Q = nC_v\Delta T = 997.68J$$ </p>
<p>I obtained 1995.36J (I used n=2).</p>
<p>To compute Q at constant pressure:</p>
<p>$$Q = nC_p\Delta T = 1662.80J$$ </p>
<p>I obtained 3325.6J (I used n=2).</p>
<p>I do not understand why it uses n=1 as it is specified that the gas is formed by 2 mols of oxygen. </p>
<p>b) When heat is supplied at constant pressure the gas disseminates and exerts a positive work (such as on a piston).</p>
<p>However the answer is given as a negative number, which means work is done by the system and not over it. Thus the given answer:</p>
<p>$$W = -nR\Delta T = -1330.24J$$ </p>
<p>Here I got the same but positive.</p>
<p>c) Using the first principle of thermodynamics:</p>
<p>$$ \Delta E_i = Q + W$$</p>
<p>When volume is constant:</p>
<p>$$ \Delta E_i = Q_v + W = -332.56J$$</p>
<p>When pressure is constant:</p>
<p>$$ \Delta E_i = Q_p + W = 332.56J$$</p>
<p>What I got before seeing the answers:</p>
<p>When volume is constant:</p>
<p>$$ \Delta E_i = Q_v + W = 3325.60J$$</p>
<p>When pressure is constant:</p>
<p>$$ \Delta E_i = Q_p + W = 4655.84J$$</p>
<p>As you can see all comes down to the definition: $Q = nC_v\Delta T$ I have made some research and in Tipler and $C_v$, $C_p$ and Q are defined: </p>
<p>$$C_v = n\frac{3}{2}R$$</p>
<p>$$C_p = n\frac{5}{2}R$$</p>
<p>$$Q = C\Delta T$$</p>
<p>Eventually it is the same, since I multiplied by 2 (number of mols) in the definition: $Q = nC\Delta T$ and I did not do it my first $C_v$ and $C_p$ definitions. Therefore I think my outcomes are right but what do you think?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2984/both-molar-specific-ideal-monatomic-constant-volume-pressureThu, 09 Aug 2018 08:59:16 +0000Mean free path of ideal gas
http://physics.qandaexchange.com/?qa=2974/mean-free-path-of-ideal-gas
<p>There's a box with a wall in mid dividing it in two sections, each filled with same ideal gas in both sections at 150 K in one section and at 300 K in another. How am I supposed to calculate ratio of mean free paths (MFP)in 2 sections?</p>
<p>My attempt: MFP ~ Volume / Number of particles</p>
<pre><code> => MFP ~ Temperature / Pressure
</code></pre>
<p>Now, Assuming pressure to be same on both sections, ratio must be half. But that is incorrect. Why?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2974/mean-free-path-of-ideal-gasTue, 07 Aug 2018 13:38:41 +0000Satellite on circular orbit around Earth
http://physics.qandaexchange.com/?qa=2933/satellite-on-circular-orbit-around-earth
<blockquote><p>A satellite of 100 kg is on equatorial circular orbit around the Earth, spinning in the same Earth direction, at a height of 1000 km over the blue planet's surface. <br>
a) How much time takes the satellite to go through the same point of the Earth’s vertical (both have rotation movement).<br>
b) What’s the total energy of the satellite on the orbit?<br>
Use: G= $6.67\times 10^{-11} \frac{Nm^2}{kg^2}$ ; Earth’s radius: $6370 km$; Earth’s mass= $5.98\times 10^{24} kg$</p>
</blockquote>
<p>What I thought:</p>
<p>a) I interpreted this question is asking for the time both satellite and Earth take to align. I came up with the idea I had to equal the two angular displacement $\theta_{1}$ and $\theta_{2}$ (satellite's one and Earth's one respectively). We have $\theta_{1} = \omega_{s}T= \theta_{2} = 2\pi + \omega_{e}T$ I also supposed Earth has always an initial angular displacement of $2\pi$. Then I just had to isolate T factor. However I did not get the given outcome $6.79\times 10^{3}s$</p>
<p>The velocity of satellite:</p>
<p>$v= \sqrt{\frac{GM}{R_{e}+h}} = 7.36\times 10^{3}m/s$</p>
<p>$\omega_{s} = \frac{v}{R_{e}+h} = 9.98\times 10^{-4} rad/s$</p>
<p>$\omega_{e} = \frac{2\pi}{T} = 7.27\times 10^{-5} rad/s$</p>
<p>$T=\frac{2\pi}{\omega_s-\omega_e} = 6.79\times 10^{3} s$</p>
<p>b) Thanks to the provided link I got the following=</p>
<p>$E = K + U = -\frac{GMm}{2(R_{e}+h)} = -2.71\times 10^{9} J$</p>
<p>I do think the provided answer $-2.71\times 10^{13}$ is wrong . It is just a factor issue. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2933/satellite-on-circular-orbit-around-earthThu, 26 Jul 2018 23:47:35 +0000Rod collides in a perfect inelastic way with a ball
http://physics.qandaexchange.com/?qa=2575/rod-collides-in-a-perfect-inelastic-way-with-a-ball
<p>Text of the problem: A uniform rod of length $L1$and mass $M$ equal to 0.75 kg is supported by a hinge of negligible mass at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass $m$ = 0.50 kg is supported by a thin string of length $L2$ from the hinge. The particle sticks to the rod on contact. What should be the ratio $ \frac {L2}{L1}$ so that $ \theta _{max} = 60°$ after the collision? (Literally copied from Tipler, chapter 10, exercise 67; last edition)<br>
<img src="https://cdn.pbrd.co/images/Hh0zsAn.jpg" alt=""><br>
Notice rod goes towards the ball</p>
<p>These are my solutions (based on an answer-pdf)</p>
<p>As we know, angular momentum is conserved due to the fact the torque exerted by the rod on the ball is equal and opposite to the torque exerted by the ball on the rod. Therefore:</p>
<p>$ \Delta L = L_{af.c} - L_{bf.c} $ </p>
<p>The system's moment of inertia DOES NOT CHANGE. It is equal to $ \frac{ML_1^2}{3} + mL_2^2 $ What I did wrong was the following : I considered the moment of inertia before the perfectly inelastic collision (PIC) equal to the moment of inertia after the PIC. The moment of inertia before the PIC is equal to the rod one and the moment of inertia after the PIC equals to the one of the system (rod+ball).</p>
<p>$ (\frac{ML_1^2}{3} + mL_2^2) \omega_{af.c} = (\frac{ML_1^2}{3} ) \omega_{bf.c} $</p>
<p>Well, as I do not know neither $ \omega_{bf.c} $ or $ \omega_{af.c} $ we have to find 2 more EQ:</p>
<p><strong>Firstly</strong>, in order to obtain $ \omega_{bf.c} $ I analysed the conservation of energy of the rod (as before the collision, the velocity of the system is only due to the rod). Therefore:</p>
<p>$ U_f - U_o + K_f - K_o = 0 $ -----> $ \frac{MgL_1}{2} - MgL_1 + \frac{I \omega_{bf.c}}{2} $</p>
<p>At this point I also had doubts related with the value of initial potential energy. At pdf is pointed out the following: the zero of gravitational potential energy be at a distance $L_1$ below the pivot, so I guessed the value was $MgL_1$ because the center of mass of the rod (initially) it is at a height equal to $L_1$.</p>
<p>Therefore</p>
<p>$$ \omega_{bf.c} = \sqrt{ \frac{3g}{L_1}} $$</p>
<p><strong>Secondly</strong>, I calculated the angular velocityof the system after the collision by conservation of angular momentum</p>
<p>$ (\frac{ML_1^2}{3} + mL_2^2) \omega_{af.c} = (\frac{ML_1^2}{3} ) \omega_{bf.c} $</p>
<p>Therefore: </p>
<p>$$ \omega_{af.c} = \frac{ML_1^2}{3(\frac{ML_1^2}{3} + mL_2^2)} \omega_{af.c}$$</p>
<p>Imagine two balls approaching to each other. When both collide stick. $v$ is their initial velocity, therefore KE of the system (ball+ball) is $ KE= 2 \frac{mv^2}{2} $ But final velocity equals to 0. Therefore KE is not conserved (I learned this example from John Rennie). However, energy is conserved before and after the collision due to the lack of friction forces.</p>
<p>Once we have reached this point, we make an energy approach (system's rotational kinetic energy is transformed to gravitational potential energy). </p>
<p>Here I do not obtain the same equation than the answer sheet, which is:</p>
<p>$\frac{-I\omega_{af.c}}{2} + \frac{MgL_{1}}{2}(1-cos \theta_{max}) + mgL_{2}(1-cos \theta_{max}) =0 $</p>
<p><strong>I realized where I got wrong. But I have difficulty determining the height at potential energy. Please could you check if the following reasonings are right?</strong></p>
<p>Taking into consideration the zero of gravitational potential energy is placed horizontally at a distance $L_{1}$ below the pivot </p>
<p>1) Gravitational potential energy of the rod at its initial position equals to $ MgL_1$. Height equals to $L_1$ due to the fact the rod's center of mass is at a distance $L_1$ from the zero of gravitational potential energy.</p>
<p>2) Gravitational potential energy of the rod at its final position (just an instant before hitting the ball) equals to $ \frac{MgL_1}{2}$. The height equals to $ \frac{L_1}{2} $ due to the fact the rod's center of mass is at a distance $ \frac{L_1}{2} $from the zero of gravitational potential energy.</p>
<p>Then my question now is: Why is there not Gravitational potential energy just after the perfectly inelastic collision happens?</p>
<p><a rel="nofollow" href="http://www.nhvweb.net/nhhs/science/rquinn/files/2013/12/Ch10-Homework-Answers.pdf">http://www.nhvweb.net/nhhs/science/rquinn/files/2013/12/Ch10-Homework-Answers.pdf</a></p>
<p>Thanks</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2575/rod-collides-in-a-perfect-inelastic-way-with-a-ballTue, 17 Apr 2018 15:09:34 +0000Resistance independent of cells ?
http://physics.qandaexchange.com/?qa=2411/resistance-independent-of-cells
<p>At what value of Resistance $R_I$ in the circuit shown in figure will the total resistance between points A and B be independent of the number of cells ? </p>
<p>I didn't understand what does question mean by asking independent of number of cells ?<br>
Rest I feel it's based on some infinite resistors like problem but don't get the last words.<img src="https://cdn.pbrd.co/images/H2KMp4w.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2411/resistance-independent-of-cellsSat, 13 Jan 2018 15:53:30 +0000Angular impulse on a pivoted rod from two different points
http://physics.qandaexchange.com/?qa=2282/angular-impulse-on-a-pivoted-rod-from-two-different-points
<p>My actual question is that <strong>does the point about which angular momentum relation is calculated need to be inertial</strong>? I mean do we have to be in the reference frame of a point to write angular momentum about that point? This is illustrated in the following question, particularly in <strong>Method 2</strong></p>
<p><strong>Q.</strong> Suppose we have a rod pivoted at the top. A linear impulse $J$ is given to rod at point B in horizontal direction. We need to calculate the linear impulse $J'$ provided by <strong>pivot</strong>. (Length of rod is $L$, mass is $m$, moment of inertia is $I = \frac{mL^2}{3}$ about pivot. Also consider that rod gets an angular velocity $\omega$ just after the impulse acts.)</p>
<p><img src="https://pasteboard.co/GMfTtKf.png" alt=""></p>
<p>I did this problem using two different approaches:</p>
<p><strong>Method 1:</strong> Angular momentum and impulse relation about point $A$ (pivot).</p>
<p>Total angular implulse (external) is $JR$ about point $A$, or the pivot. This must equal change in angular momentum, ie, $I \omega$. So we obtain $$J = \dfrac{I \omega}{ L} = \dfrac{m\omega L}{3}$$<br>
Also we know that net linear impulse on rod is equal to change in its linear momentum. we obtain $$J - J' = \dfrac{m\omega L}{2}$$</p>
<p>From these two equations we obtain $$J' = \dfrac{-m\omega L}{6}$$</p>
<p><strong>Method 2:</strong> Angular momentum about point B.</p>
<p>I started reasoning as follows: Since an impulse acts on rod at point $B$, then if we take reference point to be $B$, then we apply a pseudo force from center of mass $C$, or an impulse $J$ from $C$ in opposite direction. Now we write angular momentum - angular impulse relation about $B$.</p>
<p>$$\dfrac{JL}{2} + J'L = I \omega = \dfrac{mL^2\omega}{3}$$</p>
<p>using the value of J from above, we get:</p>
<p>$$J' = \dfrac{m\omega L}{6}$$</p>
<p>Which is clearly different from result of <strong>method 1</strong>.</p>
<hr>
<p>To me, method 1 seems fine but there seems to be a <strong>major conceptual error</strong> in method 2. Can you please explain more about angular momentum in this situation? </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2282/angular-impulse-on-a-pivoted-rod-from-two-different-pointsMon, 25 Sep 2017 22:48:15 +0000Pressure wave equation
http://physics.qandaexchange.com/?qa=2204/pressure-wave-equation
<p><img src="http://dc.allenbpms.in/testpaper/solution/d10245-20-636302088603853855-020s.gif" alt=""></p>
<p>In the first equation why amplitude is $2P_0$ .<br>
And I think the equation is wrong . </p>
<p><a rel="nofollow" href="http://byjus.com/physics/longitudinal-waves/">the equation should be</a></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2204/pressure-wave-equationSun, 14 May 2017 16:43:08 +0000Potential at different plate
http://physics.qandaexchange.com/?qa=1751/potential-at-different-plate
<p><img src="http://dc.allenbpms.in/testpaper/solution/d8527-11-636195703771577709-011s.gif" alt=""></p>
<p>In the solution they have done $V_2 -V_1 = V_2 -V_3$</p>
<p>But I want to know can we solve it by taking $$V_1 -V_3=0$$ $$V_1 = V_3$$<br>
from that we will get $$\frac{-x(3d)}{A\epsilon}= \frac{(x-5Q)(3d)}{A\epsilon}$$ $$ -x=5Q+x$$<br>
But using that I am not getting the answer .</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1751/potential-at-different-plateMon, 27 Mar 2017 10:16:23 +0000voltage gain of npn transistor
http://physics.qandaexchange.com/?qa=1749/voltage-gain-of-npn-transistor
<p><img src="http://dc.allenbpms.in/testpaper/solution/d8527-29-636195703771799106-029s.gif" alt=""></p>
<p>As V$_{in}$ increases V(out) also increases.<br>
Then how can we say Av is negative and its magnitude is increasing'</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1749/voltage-gain-of-npn-transistorMon, 27 Mar 2017 09:31:50 +0000Maximum tension ia the rod
http://physics.qandaexchange.com/?qa=1710/maximum-tension-ia-the-rod
<p><img src="http://dc.allenbpms.in/testpaper/solution/d9336-18-636260591738978628-018s.gif" alt=""></p>
<p>In this I could not understand how they have got the lengths as 2l/3 and l/3 .<br>
Is there any trick behind this or error in the question.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1710/maximum-tension-ia-the-rodSun, 26 Mar 2017 09:56:53 +0000Find relative refractive index when Brewster's angle is given
http://physics.qandaexchange.com/?qa=1483/find-relative-refractive-index-when-brewsters-angle-given
<p>A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $i_C$ and the Brewster's angle of incidence is $i_B$, such that $sini_C/sini_B = = 1.28.$<br>
The relative refractive index of the two media is </p>
<p>I know $tan i_B=\mu $(refractive index)<br>
And in this case sin$i_C = \mu$<br>
So putting these values in$sini_C/sini_B = = 1.28.$<br>
I got $\sqrt{1+\mu^2}=1.28$<br>
From this I got approx $\mu =0.6$<br>
But answer is 0.8</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1483/find-relative-refractive-index-when-brewsters-angle-givenSat, 04 Mar 2017 15:44:47 +0000Number of ideal gas molecules per $cm^3$
http://physics.qandaexchange.com/?qa=1461/number-of-ideal-gas-molecules-per-%24cm-3%24
<p>The lowest pressure that is the best vacuum that can be created in a laboratory at 27 degree celsius is $10^{-11}$millimetre of Mercury at this pressure the number of ideal gas molecules per $cm^3$ will be ?</p>
<p>As said ideal gas so I applied Ideal gas equation<br>
$ PV=nRT $<br>
$[(10^{-11}×10^5)/(24.4×760)]×10^6$<br>
this should give $n/V $ <br>
and multiplying it by avogadro number ahould give answer but it isn't ?</p>
<p>Symbols have there usual meanings </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1461/number-of-ideal-gas-molecules-per-%24cm-3%24Fri, 03 Mar 2017 17:17:12 +0000when will the site be completed?- part 2
http://physics.qandaexchange.com/?qa=1447/when-will-the-site-be-completed-part-2
<p>On 7 December 2016 physicsapproval asked <a rel="nofollow" href="http://physicsproblems.nfshost.com//?qa=467/">When will the site be completed and up on the search engines?</a> The site was then not much more than 1 month old. Almost 3 months later we are still in the same position. </p>
<p>Have the founders and developers abandoned the site? What progress is being made? What is preventing the site from being available to search engines?</p>
Metahttp://physics.qandaexchange.com/?qa=1447/when-will-the-site-be-completed-part-2Thu, 02 Mar 2017 08:36:07 +0000Find the position of the lens
http://physics.qandaexchange.com/?qa=1102/find-the-position-of-the-lens
<p>In the given optical instrument in which manner third lens of same focal length can be inserted in order to increase the collecting efficiency of light without changing the position of object and image. calculate the position of that lens<img src="https://cdn.pbrd.co/images/so2lrwOjS.jpg" alt=""><br>
I am unable to understant collecting efficiency I tried to search it on Internet but still didn't get much help , then how should I proceed ?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1102/find-the-position-of-the-lensMon, 30 Jan 2017 18:00:59 +0000Photocell in the circuit
http://physics.qandaexchange.com/?qa=1014/photocell-in-the-circuit
<p>A bulb B having a resistance of 100 ohm converts 31 % of its electrical energy into radiation having wavelength less than 5000 A and with an average wavelength of 4000 A in this range. 10% of this radiation is incident on a photocell D of efficiency 20% and is converted into an electric current. The photo electric threshold of the material is 5000 A. <br>
The photocell (D) is connected in parallel with a resistance R (100 ohm) and a capacitance C (=100 $\mu$F) and finally in series with the bulb B & a source of emf E = 180 V. The key is now closed.<br>
<img src="http://i.imgur.com/wJEIjCY.jpg" alt=""><br>
Now we have to find the charge on C and power dissipiated in (both in steady state)</p>
<p>I am not getting any start, now I am able to understand the given <a rel="nofollow" href="http://dc.allenbpms.in/testpaper/solution/d8679-29-636206218321103018-029s.gif">solution</a> in my book .</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1014/photocell-in-the-circuitTue, 24 Jan 2017 17:17:37 +0000Charge deposited on the capacitor by light
http://physics.qandaexchange.com/?qa=1013/charge-deposited-on-the-capacitor-by-light
<blockquote><p>Light consisting of wavelengths 4000 A° & 6000 A° is incident on a double slit of separation 0.1mm kept at a distance of Im from a screen.<br>
A small rectangular photodetector, 2mm x 10 mm in size, of efficiency 20% and work function 2.2 eV is placed lengthwise. parallel to the fringes. The photoelectrons constituting the current are swept off and the photo current (ip) is allowed to pass through the circuit shown in the diagram.<br>
The photocurrent i$_p$,given that intensity of light at the maximum of the central fringe is 155W/m$^2$ due to each wavelength and the photodetector is placed at a position where the current i$_p$ is maximum. Assume that all photoelectrons have the same energy and that the extemal voltage is sufficient to sweep all the photoelectrons to form the current.<br>
We have to find the charge deposited on the capacitor C ( =120 u F) in 2 secs after the source of light (of both wavelengths) is switched on.<img src="http://i.imgur.com/XVVx1sF.jpg" alt=""></p>
</blockquote>
<p><strong>I tried as</strong><br>
<img src="http://i.imgur.com/yXCPEJ2.jpg" alt=""><br>
<strong>Now how can I proceed.</strong></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=1013/charge-deposited-on-the-capacitor-by-lightTue, 24 Jan 2017 16:37:52 +0000Private message
http://physics.qandaexchange.com/?qa=981/private-message
<p>Hello!</p>
<p>Why does the website tell me that I need 150 reputation to access private messages, even though I have 168 points as of now?</p>
<p><img src="http://i.stack.imgur.com/OfTpx.png" alt=""></p>
Metahttp://physics.qandaexchange.com/?qa=981/private-messageTue, 24 Jan 2017 01:32:14 +0000Negative velocity squared ?
http://physics.qandaexchange.com/?qa=979/negative-velocity-squared
<p>>A Venturi tube without manometer is given with initial area $A$ and area in its throat as $a$, initial pressure as $P_1 = 3 \ \ \mathrm{atm}$ and $A = 5a$, Find velocity of water as it enter the tube.</p>
<p><img src="https://cdn.pbrd.co/images/pcJD3VLcM.png" alt="Figure"> </p>
<p>by equation of continuity :- </p>
<p>$$Av_1 = av_2 \implies 5av_1 = av_2 \implies 5v_1 = v_2 \tag{1}$$</p>
<p>Since $\Delta F = 0$<br>
Thus, $$P \propto \frac1A \implies P_1 A = P_2a \implies P_1\times 5a = P_2a \implies 5P_1 = P_2 \tag{2}$$</p>
<p>By Bernoulli's principle :-</p>
<p>$$\Delta P + \rho g\Delta h + \frac12\rho\Delta v^2 = 0$$<br>
$$(P_2 - P_1) + \frac12\rho(v_2^2 - v_1^2) = 0$$</p>
<p>From 1 and 2, </p>
<p>$$(5P_1 - P_1) + \frac12\rho(25v_1^2 - v_1^2) = 0$$<br>
$$(P_1) + \rho(3v_1^2) = 0$$</p>
<p>$$\sqrt{-\frac{P_1} {3\rho }} = v_1$$</p>
<p>Now this is impossible and I am a idiot to get to this result but even after staring at this for, I can't get spot the flaw. Surely some of you can do that.</p>
<p>I am grateful for any help . </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=979/negative-velocity-squaredSun, 22 Jan 2017 22:50:29 +0000When will the site be completed?
http://physics.qandaexchange.com/?qa=467/when-will-the-site-be-completed
<p>I just wanted to know when the site would be up on the search engines. Is any help, like adding more variety of questions related to different tags, from the user side required?</p>
Metahttp://physics.qandaexchange.com/?qa=467/when-will-the-site-be-completedWed, 07 Dec 2016 17:24:37 +0000Moderator Guidelines
http://physics.qandaexchange.com/?qa=244/moderator-guidelines
<p>Below is a start of a list of moderation guidelines or principles. Over time we can expand this list and increase its sophistication.</p>
<ul>
<li><p><strong>Spam and Offensive Content</strong> - A moderator should hide a question, comment or answer immediately if the content is offensive, spam or contains copyright material.</p>
</li>
<li><p><strong>Closing Questions</strong> - As the site is in its infancy, I think we should have at least <strong>two</strong> moderators agree before we close a question as being off topic. This is because closing for off-topic can often be a subjective decision. (While we develop a close voting system, please flag proposed questions to close in the moderator chat)</p>
</li>
<li><p><strong>Editing Posts</strong> - Moderators may edit posts to improve quality (e.g. correcting grammatical mistakes or adding Latex to a post). Moderators should not make substantial changes to a posts content that results in the post differing from the original users intention.</p>
</li>
<li><p><strong>Be Friendly and Welcoming to New Users</strong> - New users are often unfamiliar with the Q&A model as well as the standards we expect. Please bear this in mind when communicating with new users. </p>
</li>
<li><p><strong>Participate</strong> - Politely and in accordance with site rules.</p>
</li>
<li><p><strong>Banning Users</strong> - Should only be done for serious offences (which violate rules that need be clearly defined in another post) and must always be discussed with other moderators.</p>
</li>
<li><p><strong>Discuss Decisions with Other Moderators</strong> - the bigger the decision, the more discussion there should be.</p>
</li>
</ul>
Metahttp://physics.qandaexchange.com/?qa=244/moderator-guidelinesTue, 08 Nov 2016 12:46:28 +0000Under Construction - What's left?
http://physics.qandaexchange.com/?qa=168/under-construction-whats-left
<p>This site is currently still under construction (shout out to Einstein here, for building this amazing site from scratch). This is a list of things that we haven't yet added to the site, just to get things started. If you have things to add that I forgot, please don't answer, but edit the question and add them. Hopefully this will be one single to-do list that will contain everything we have to do.</p>
<h3>Technical</h3>
<ul>
<li><strong>LaTeX not rendering properly in several cases. See <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=180/lorentz-invariant-integration-measure&show=181#a181">this answer</a>, <a rel="nofollow" href="http://physicsproblems.nfshost.com//?qa=191/work-and-kinetic-energy-on-incline">this question</a>, and <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=179/a-question-based-on-galilean-relativity">this question</a>.</strong></li>
<li>Close vote and flagging reasons (as described in the <a rel="nofollow" href="http://physicsproblems.nfshost.com//?qa=128/close-reasons">close votes meta post</a>)</li>
<li><span style="text-decoration: line-through;">The Q&A and Questions tabs seem to have the same things in them</span>
</li>
<li>Not many tags - maybe import those from Physics.SE?</li>
<li>Full chat room, for moderators and non-moderators, with functionality like that of the SE chat</li>
<li>Meta and normal site separation (easier to switch between the two)</li>
<li>Remove "banned" users tab</li>
<li><span style="text-decoration: line-through;">Auto-fit of image size (See [this question](http://physicsproblems.nfshost.com/?qa=145/moment-of-force-exerted-by-water-on-tube) where the uploaded image has been literally cropped out while being uploaded) </span>
</li>
<li>Currently while editing any previous answer there is a constant shaking motion in the preview (with a message at the left bottom saying "Processing Math") which seems quite distracting. We need to stabilize the preview (like in the Stack Exchange sites).</li>
<li>Pinging Users: Currently if we reply to a certain message all the people who had participated in the comment thread get notified. Single user pinging should be activated like in the Stack Exchange sites. (Maybe multi-user pinging too?) </li>
<li>Optimization for mobile browsers like in Android or IOS. </li>
<li>Removing the accounts of bots.</li>
<li>Edit system: Whom should we allow to edit posts? Should there be peer review system of edits before an user reaches a certain reputation? This is something to discuss.</li>
<li>Close vote/review system - systems for this like on SE; this goes with the previous one.</li>
<li>Maybe an Android and IOS app too? (This should be placed last in order of priority.)</li>
<li>It seems to be impossible to delete comments, whether you are a moderator or not. This would be a good thing to add (deletion of your own comments for non-moderators, deletion of any comments for moderators).</li>
<li>Flagging comments is not yet existent. This would be reasons such as obsolete, spam, not constructive, too chatty, etc, like the ones on SE.</li>
<li>Moving comments to chat - this obviously depends on having chat.</li>
<li><span style="text-decoration: line-through;">Make acceptance of an answer allowed only by the user who asked the question.</span>
</li>
<li><span style="text-decoration: line-through;">Weird problem - the notifications tab is not showing up centered for one user. It is occurring using chrome on a windows 10 operating system:</span>
</li>
<li>when adding an image to a question/answer, you are unable to upload the image from your computer</li>
<li>Upvote/downvote/answer/comment notifications seem to be showing up twice in the feed.</li>
<li>Formatting/in-text links not working in comments</li>
</ul>
<h3>Non-Technical</h3>
<ul>
<li><span style="text-decoration: line-through;">Logo</span>
</li>
<li><span style="text-decoration: line-through;">Background/site design </span>
</li>
<li><span style="text-decoration: line-through;">The current design i.e. white screen on a large grey background should be replaced with a single background. The screen space is getting wasted currently on the extra large grey borders </span>
</li>
<li><span style="text-decoration: line-through;">Site name</span>
</li>
<li><span style="text-decoration: line-through;">Legalese - terms and conditions</span>
</li>
<li>Debate over the rep system - what are appropriate amounts? Are there any changes we need to make?</li>
<li>Advertising the site on Google and using Search Engine Optimization plugins.</li>
</ul>
<p>Again, please edit the question and add any things you think of.</p>
Metahttp://physics.qandaexchange.com/?qa=168/under-construction-whats-leftFri, 04 Nov 2016 11:55:22 +0000