Recent content by haxxorboi
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Calculating U-235 Requirements for a 707 MW Nuclear Fission Reactor
Holy crap, THANK YOU! 22.29595E15 joules heat energy per year at 707 MW 2.9464E-11 Joules per event 2.5617E24 atoms per kg 75.478E12 Joules per kg and then obviously divide to get final answer of kg. That one was killing me, thanks so much both of you.- haxxorboi
- Post #4
- Forum: Introductory Physics Homework Help
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Calculating U-235 Requirements for a 707 MW Nuclear Fission Reactor
Homework Statement How many kilograms of U-235 would be needed to run a 707 MW reactor for 1 year? Homework Equations 183.9 MeV released per reaction 7.57E26 reactions per year (These numbers are from the prior problems) The Attempt at a Solution I thought maybe half life, but...- haxxorboi
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- Fission Nuclear Nuclear fission Reactor
- Replies: 3
- Forum: Introductory Physics Homework Help
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Focusing on Object K: Magnification & Adjustment of Focal Length
*sigh* Yep, that'd be the issue... need either more sleep or more caffeine I guess... Problem solved. Thank you very much... I'm going to go take a nap...- haxxorboi
- Post #3
- Forum: Introductory Physics Homework Help
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Focusing on Object K: Magnification & Adjustment of Focal Length
Homework Statement Suppose the lens in a human eye has far focal length equal to f0 = 40.1 mm, for distant objects to be in focus. A. An object K is 1.69 m from the lens. How much should the focal length change to focus on the object K? B.What is the magnification ( = image size...- haxxorboi
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- Focal Focal length Length Magnification
- Replies: 2
- Forum: Introductory Physics Homework Help
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Solve Snell's Law Problem: Determine θ
Man, what the heck was I doing? I did it again and I got the same answer as you, which is the correct answer... I swear I just do things wrong 10 times in a row for no reason... Thanks for the help!- haxxorboi
- Post #3
- Forum: Introductory Physics Homework Help
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Solve Snell's Law Problem: Determine θ
Homework Statement A fish is 30 cm below the surface of a lake. If he looks up he can see the entire sky compressed in a cone of angle θ, because of refraction of light at the water surface. (θ is the angle between the cone axis and the cone surface.) Determine θ. Homework Equations...- haxxorboi
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- Law Snell's law
- Replies: 3
- Forum: Introductory Physics Homework Help
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Understanding Flux and EMF in a Rotating Conducting Coil: True or False?
Anyone? I'm struggling here bad...- haxxorboi
- Post #3
- Forum: Introductory Physics Homework Help
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Understanding Flux and EMF in a Rotating Conducting Coil: True or False?
Ok, upon further thinking I've discovered in this case that... Flux=sin(x) EMF=-cos(x) 1. In B the flux through the coil is zero and increasing. 2. In A the flux through the coil is zero and decreasing. 3. In B the flux through the coil has the maximum positive value. 4. In A the emf...- haxxorboi
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- Forum: Introductory Physics Homework Help
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Understanding Flux and EMF in a Rotating Conducting Coil: True or False?
Homework Statement The conducting coil rotates around the axis of rotation in the direction of the right-hand rule (thumb pointing along the axis of rotation, the fingers curl in the direction of rotation). A uniform magnetic field B points from left to right. The "flux" refers to the...- haxxorboi
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- Generator Properties
- Replies: 2
- Forum: Introductory Physics Homework Help
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Magnetron Electron Field Strength
Velocity=Distance/Time Velocity=Circumference/Period Velocity=2*Pi*R/6.8E-11 sec Velocity=9.24E10R 1.6E-19*B=9.1E-31*(9.24E10R)/R B=.5253 T Thank you very much, that was killing me!- haxxorboi
- Post #3
- Forum: Introductory Physics Homework Help
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Magnetron Electron Field Strength
Homework Statement Suppose the resonant frequency is f = 1.47×1010 s−1; that is, the electron period of revolution is T = 6.80×10-11 s. What is the corresponding strength of the magnetic field? Homework Equations qvb=mv2/r qb=mv/r The Attempt at a Solution 1.6E-19*b=9.1E-31*v/r...- haxxorboi
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- Electron Field Field strength Strength
- Replies: 2
- Forum: Introductory Physics Homework Help
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How to Calculate Remaining Charge on a Capacitor After Discharge?
SWEET! The answer ended up being .5591 C which was correct with the system. Thanks so much for the help guys, I feel stupid with not recognizing which "e" that was seeing as this is an exponential decay. I'm not sure how we were supposed to know how to do this problem though as I can't even...- haxxorboi
- Post #6
- Forum: Introductory Physics Homework Help
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How to Calculate Remaining Charge on a Capacitor After Discharge?
I've never seen that equation before and I can't find it in our chapter. Let me see if I understand this equation... Final charge=1 Coulomb * 1.602E-19-.004/(86*8E-5) Does that seem right or am I misinterpreting part of the equation. Thanks- haxxorboi
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- Forum: Introductory Physics Homework Help
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How to Calculate Remaining Charge on a Capacitor After Discharge?
Homework Statement A capacitor is charged to 1 coulomb; the capacitance is 8.00×10-5 farads. Then a switch is closed which puts the capacitor in a closed circuit with a resistor; the resistance is 8.60×101 ohms. Calculate the charge on the capacitor after 4.00 milliseconds. (1 ms = 0.001 s)...- haxxorboi
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- Capacitor Discharge
- Replies: 6
- Forum: Introductory Physics Homework Help
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Theoretical power from wind turbine
So your rho is approximately 1.1215 kg/m3? I'm guessing this is not counting Betz limit as this is a Phy101 class. What frustrates the hell out of me is we haven't discussed this at all and it's not even in the readings. We're doing the buoyancy right now, which IS fluids I suppose, but it's...- haxxorboi
- Post #6
- Forum: Introductory Physics Homework Help