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Latest comments
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| Latest 25 Comments | | 2 | Click here to jump to the problem! | gfsampaio
2024-03-12 20:47:37 | I make for onde form, i get the solution of Schrödinger equation time independent, were the wave function is give for Ψ = A exp(- (2mV(x)/ ℏ²)^1/2 * x), this have to be equal to the wave function given by the question, so (2mV(x)/ ℏ²)^1/2 * x = b²x²/2 resulting in V(x) = ℏ²(b^4)x²/8M. I make problem with this 8M, i don't understand for physical interpretation of him, or if i make any basic error in my solution. I see too some answers with the choice E, this is impossible, if E is the right answer, the potential in x=0 is different of 0. |
| 5 | Click here to jump to the problem! | dotyhughes117
2023-07-05 19:51:20 | I noticed that you’ll be are usually legitimately interested in this type of! We are seeking to build my own ring online site and you also include reduced the problem by including great statistics.
https://www.sydneyborewater.com.au/ |
| 6 | Click here to jump to the problem! | chavesarlene4
2023-05-10 06:09:37 | Electric charges, magnetic moments, and the electromagnetic field physically interact to form electromagnetism. A wave can form in the electromagnetic field, or it might be static or slowly changing. Light generally refers to electromagnetic waves that follow the principles of optics Grande Prairie Snow Removal |
| 9 | Click here to jump to the problem! | emmo
2019-12-18 06:28:15 | According to this solution, "the cross-product yields 0 force for the two horizontal components," but is that right? The field is pointing into the page, which is orthogonal to the line of the induced current, thus there would be force on each horizontal component. This contributes nothing to the net force due to symmetry: the top and bottom horizontal portions of the induced current cancel one another out. |
| 10 | Click here to jump to the problem! | alwayswright5214
2019-12-02 13:39:56 | Compton effect (Compton, 1923), named after its discoverer, the greatest photon energy loss occurs when it is scattered backward (180°) from its original direction. Then, if E is much larger than the rest energy of the electron E0 = me c2 = 0.511 MeV, it is found that the final photon energy E′ is equal to E0/2. www.dc-harvest.com/ |
| 12 | Click here to jump to the problem! | toto212
2019-11-16 05:06:24 | Hello , i Just want to say that, umm..\\\\r\\\\nwhat u make is very amazing and i also like it ^.^\\\\r\\\\nThanks friend. I hope i can see your various works.\\\\r\\\\nKeep working guys. jayatogel |
| 14 | Click here to jump to the problem! | droosenoose
2019-10-22 23:07:47 | this is a bit misleading because you can easily use negative feedback to amplify signal if you have an op amp and two resistors, which is what I immediately thought of when they said negative feedback.
Guess they were referring to the concept of negative feedback itself, not applications?? |
| 17 | Click here to jump to the problem! | PHYSGOGO
2019-10-22 04:40:35 | oh sorry guys, I found another way: The energy to remove 2 electrons(79.0)= remove 1st+remove 2nd. Since remove 1st is under help of second electron's repel force. Removing second is 2^2*13.6=54.4 remove 1st=79-54.4=24.6 |
| 18 | Click here to jump to the problem! | PHYSGOGO
2019-10-22 04:34:15 | since there are interaction between two electrons. we can solve in this way: STEP1： Remove two electrons(79eV) STEP2: pull one electron back(13.6*2^2=54.4eV) the result is also equals: ionize one electron. \\\\r\\\\nThe trick is, we can just calculate ionize energy of isolated electron, so we have to find a way isolate them. \\\\r\\\\nThe result is: 79-54.4=24.6 (we set remove is plus,pull back is minus) |
| 20 | Click here to jump to the problem! | Ryry013
2019-10-19 12:36:04 | Man this site is buggy. Looked alright in the preview! Basically, it was V is proportional to dB/dt which is about equal to ΔB/Δt. There's 3 ΔB per revolution (as in, min-->max, changing from N-->S), so if we multiply (10 rotations/second)(3 ΔB/rotation) we get 30 ΔB/second |
| 24 | Click here to jump to the problem! | droosenoose
2019-10-18 00:46:50 | Consider conservation of energy. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\r\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\n\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\r\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\nIf the block slides at a constant rate down the wedges a distance d then the potential energy lost is Mgd which goes to the friction (2m+M)ug*x but since 45 degree wedges x=d and you get M = 2um/(1-u) |
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