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| bio | website | inpursuitoflaziness.blogspot.… |
|---|---|---|
| location | Mumbai, India | |
| age | 19 | |
| visits | member for | 1 year, 3 months |
| seen | 4 hours ago | |
| stats | profile views | 2,179 |
I am an engineering student who loves the sciences, especially physics.
Help this interesting site proposal for Space Exploration reach beta!
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Mar 13 |
answered | Friction + Bouncing of an Object against an Elastic Wall |
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Mar 13 |
comment |
Friction + Bouncing of an Object against an Elastic Wall You may want to check out gamedev.stackexchange.com |
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Mar 13 |
answered | Can anyone solve this simple static equilibrium problem? |
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Mar 13 |
revised |
Is the free electron wavefunction stable? edited tags |
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Mar 13 |
answered | Piston movements in four stroke cycle? |
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Mar 13 |
comment |
Resistor circuit that isn't parallel or series @RonMaimon now why didn't I think of that? Yeah, the continued fraction one looks easy to do, but the diagram may be slightly complicated.. |
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Mar 13 |
revised |
Factors affecting torque and RPM of a motor edited tags |
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Mar 13 |
comment |
Tension on a cable in a gravitational field This makes no sense in the context of the problem.. Have you read it properly? |
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Mar 13 |
comment |
Black body balloon in vacuum @EricMercer: It need not be stated, you sort of have to figure that out.. Yeah, it gets annoying when problems specify additional constraints like "pressure constant" which lead to energy loss since something has to work to keep it constant. |
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Mar 13 |
comment |
Black body balloon in vacuum Yean, $nC_VdT seems to give the answer. |
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Mar 13 |
comment |
Black body balloon in vacuum Hmmm.. It looks like you're supposed to use $dU=nC_VdT$. Which means that there is some energy being lost in keeping the internal pressure constant. For a monoatomic gas, $C_v=\frac{3R}{2}$ |
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Mar 13 |
comment |
Black body balloon in vacuum @EricMercer: Pressure is 0. It's external pressure we're talking about (we don't care about the work done by balloon on system, we care about the energy lost when it expands *against* external pressure ). Anyways, $dU=TdS-PdV$ won't get you much. You need to calculate entropy. If you try, you basically get $dQ=TdS; \: dU=dQ-dw$. Back to the FLTD. |
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Mar 13 |
comment |
Black body balloon in vacuum If it helps, the $Q$ used in the first law of thermoD is $-Q_b$ here (since I took energy emitted) |
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Mar 13 |
comment |
Black body balloon in vacuum I guess what you did for b is correct. $dQ/dt$ is the amount of energy radiated per second. For c, since $Q=U$, their corresponding rates are the same as well (there may be a $\pm$, though). |
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Mar 13 |
comment |
Black body balloon in vacuum @EricMercer: By first law of thermodynamics, $Q=U+w$. $Q$ is heat, $U$ is internal energy. $w$ is work, but it is zero here because its a vacuum. So here, heat lost=decrease in internal energy. So they're the same here. |
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Mar 13 |
revised |
Black body balloon in vacuum LaTeXify math |
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Mar 13 |
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Black body balloon in vacuum Just out of interest, Is $J$ some sort of heat-current-density? It looks like emissive power to me.. |
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Mar 13 |
comment |
When should we use the concept of Impulse/Momentum instead of Force? @w4j3d =P By 'hanging', I didn't mean he was executed (I'm quite sure that dead men catch no objects, even if in equilibrium).. I meant he was hanging-from-a-rope-like-tarzan sort of thing. If you want to avoid that, just make him sit in a large scale pan... Hypothetical situations are always strange. |
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Mar 13 |
answered | Black body balloon in vacuum |
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Mar 13 |
revised |
Good references for specific heats edited tags |
