Recent content by Saibot
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Calculate the resistance to add in parallel to balance this bridge circuit
Thanks for your input everyone!- Saibot
- Post #7
- Forum: Introductory Physics Homework Help
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Calculate the resistance to add in parallel to balance this bridge circuit
Got it, thank you. The current through the top is larger than the current through the bottom. Man, it seems so obvious now.- Saibot
- Post #6
- Forum: Introductory Physics Homework Help
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Calculate the resistance to add in parallel to balance this bridge circuit
The reading on the voltmeter is 0. I figure the potential drop through line AB must be the same as the potential drop through AC (since voltmeter is zero). As such, the 10 Ohm resistor has the same as the effective resistance of the parallel branch. 1/10 = 1/8 + 1/R 1/R = -0.025 R = -40 Ohms...- Saibot
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- Physics Resistance Voltage
- Replies: 6
- Forum: Introductory Physics Homework Help
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Converting Energy to Mass: Understanding the Relationship Between MeV and kg
Understood. Thanks so much :)- Saibot
- Post #9
- Forum: Introductory Physics Homework Help
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Converting Energy to Mass: Understanding the Relationship Between MeV and kg
OK, so it was incorrect to replace the "unit" c with the actual "value" of c (3*10^8). Got it, thanks mate.- Saibot
- Post #5
- Forum: Introductory Physics Homework Help
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Converting Energy to Mass: Understanding the Relationship Between MeV and kg
Indeed, but if I directly convert 1672 MeV/c^2 using the fundamental charge: (1672 * 10^6 eV/c^2) * 1.609 J/eV, I get 2.68 * 10^-10 J/c^2. This is kilograms. Same wrong answer. I'm missing something here. I have to divide again by c^2 and I have no idea why.- Saibot
- Post #3
- Forum: Introductory Physics Homework Help
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Converting Energy to Mass: Understanding the Relationship Between MeV and kg
(1672 MeV/c^2) * c^2 = 1.505 * 10^20 MeV = 1.505 * 10^26 eV = 2.41 * 10^7 J Since E = mc^2, m = E/c^2 Therefore, m = 2.41 * 10^7 / (3 * 10^8)^2 = 2.68 * 10^-10 kg But the answer is 2.97 * 10^-27 kg Help! What is wrong with my logic?- Saibot
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- Error Homework Introductory physics
- Replies: 8
- Forum: Introductory Physics Homework Help
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Speed of Message Relative to a Space Station: Relativistic Addition
Indeed, the message is sent via another rocket of some sort (I made the question up). It's still nice to see your thinking behind a carrier wave though! Thanks to both of you for taking the time to explain.- Saibot
- Post #6
- Forum: Introductory Physics Homework Help
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Speed of Message Relative to a Space Station: Relativistic Addition
Thanks vela :) I'm trying to self-teach this; you have no idea how much that reassurance helps.- Saibot
- Post #3
- Forum: Introductory Physics Homework Help
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Speed of Message Relative to a Space Station: Relativistic Addition
u is the speed of the message relative to the station v is the speed of the spaceship relative to the station u’ is the speed of the message relative to the spaceship u=(-0.5c+0.7c)/(1+((-0.5c)(0.7c))/c^2 ) =0.2c/0.65=0.308c This just seems way too high, and I'm not sure if I'm doing it...- Saibot
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- Relativity Spaceship Velocity
- Replies: 7
- Forum: Introductory Physics Homework Help
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Wolfram gave me one answer, examiners gave me another
Got it, thanks guys. So I'm not actually wrong, we'll both just get different expressions for C.- Saibot
- Post #4
- Forum: Calculus and Beyond Homework Help
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Wolfram gave me one answer, examiners gave me another
I removed the coefficient of 1/2 before integrating. So I had: 1/2 integral[(1/x-1)] = 1/2 ln(x-1) + C Using u substitution without removing the coefficient yields 1/2 ln(2x-2) + C I get how it works both ways, and my answer must be wrong, but what exactly is causing this conflict? Am I not...- Saibot
- Thread
- Replies: 4
- Forum: Calculus and Beyond Homework Help