Recent content by BoanviaFx
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Does Young's Modulus Change When Nano-Tubes Are Connected End-to-End?
Thanks :) !- BoanviaFx
- Post #3
- Forum: Introductory Physics Homework Help
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Does Young's Modulus Change When Nano-Tubes Are Connected End-to-End?
Homework Statement A nano-tube having cross sectional area A and length L extends ΔL when a force F is applied normally to one of the cross-sectional surfaces. Show that if N tubes are connected end-to-end the then Young's modulus of the fibre does not change. Homework EquationsThe Attempt at...- BoanviaFx
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- Modulus Young's modulus
- Replies: 2
- Forum: Introductory Physics Homework Help
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GPS Altitude Mystery at Sea Level - Solved?
Thanks for such an informative post! Pressure sensors do seem ideal for aviation rather than gps.- BoanviaFx
- Post #10
- Forum: Electrical Engineering
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GPS Altitude Mystery at Sea Level - Solved?
Yes I guess when it comes to my self guided quad-copter I'm going to have to set the initial altitude to the reference or maybe use an external altitude sensor.- BoanviaFx
- Post #7
- Forum: Electrical Engineering
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GPS Altitude Mystery at Sea Level - Solved?
russ_watters, that link you just sent me explains a lot. Hence the readings I obtained. Thank you for your help! And sorry about that daven my bad!- BoanviaFx
- Post #5
- Forum: Electrical Engineering
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GPS Altitude Mystery at Sea Level - Solved?
I'm a programmer/electronics student and I recently bought a GPS which I hooked up serially with my laptop. I didn't get into much detail of encoding the string which is output from the TX/RX pins of the GPS so I downloaded, a ready made program called "ublox 6". Longitude and latitude are...- BoanviaFx
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- Altitude Gps Mystery Sea level
- Replies: 9
- Forum: Electrical Engineering
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Thermodynamics First Law Question
Thank you so much! That helps a lot. :)- BoanviaFx
- Post #5
- Forum: Introductory Physics Homework Help
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Thermodynamics First Law Question
Thanks for your reply! Alright I see what you mean. Second attempt: Number of moles in gas using ideal gas equation: R= 8.314 4621 J mol-1 K-1 (Given in exam paper booklet): n=(PV/RT) n=(1.5x105Pa*1.25x10-3m3)/(8.314*300K) n=0.075 Change in temperature using Charles law would be: 372K...- BoanviaFx
- Post #3
- Forum: Introductory Physics Homework Help
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Thermodynamics First Law Question
Homework Statement Can someone help me confirm if I answered correctly please? a) A sample of gas is enclosed in a cylinder by a piston. The cylinder is given 225J of energy which expands and pushes the piston 16cm outwards against an atmospheric pressure of 1.01x105Pa. i) give the equation...- BoanviaFx
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- First law Law Thermodynamics Thermodynamics first law
- Replies: 4
- Forum: Introductory Physics Homework Help
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Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
Sorry lagg. Ignore this post.- BoanviaFx
- Post #22
- Forum: Introductory Physics Homework Help
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Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
(Heat lost changing state + Heat lost due to change in temperature) = Heat Gained When changing water to steam it would be heat required. But when steam is changed back to water, is it heat energy gained or lost?- BoanviaFx
- Post #21
- Forum: Introductory Physics Homework Help
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Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
Oh alright this makes more sense now. So the mass is:8.6x10-3kg- BoanviaFx
- Post #20
- Forum: Introductory Physics Homework Help
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Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
I'm confused, now I get the idea that its written like this? (ml+cΔT)=mc∆θ- BoanviaFx
- Post #18
- Forum: Introductory Physics Homework Help
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Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
Oh right. mc∆θ=mc∆θ m*4200*2.26x106Jkg-1*(100-80)=0.08*4200*(60) m=1.06x10-7kg Q=ml is the energy used to convert the water into steam and vice versa without a change in temperature, and the specific heat capacity * change in temperature indicates the heat lost?- BoanviaFx
- Post #16
- Forum: Introductory Physics Homework Help
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Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
Like this? Sorry if I might be frustrating you, I don't blame you. Left side is the water added to the cup, assuming the water is at 100C. mc∆θ=mc∆θ m*4200(100-80)=0.08*4200*(60) m=0.24kg- BoanviaFx
- Post #14
- Forum: Introductory Physics Homework Help