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*Various Questions: EPE Proton, Sound Intensity, Sound Frequency, Heat Conduction,... |
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| Mar11-10, 01:35 PM | #1 |
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*Various Questions: EPE Proton, Sound Intensity, Sound Frequency, Heat Conduction,...
Hi There,
I have a handful of questions I need help with and figured compiling them in one thread was the best option. For the most part, i've added these because I just don't know where to start and can't figure it out at all. Here we go! QUESTION ONE: 1. The problem statement, all variables and given/known data A bar of copper with length 2.635m and a bar of Aluminum with length 2.628m are sitting at room temperature, T1 = 25C. At what temperature, T2, will the two have the same length? The coefficient of thermal expansion of copper is 1.60x10-7 K-1 and of Aluminum is 2.25x10-7 K-1. 2. Relevant equations Linear Thermal Expansion: [tex]\Delta[/tex]L=[tex]\alpha[/tex]Lo[tex]\Delta[/tex]T Lo = initial Length 3. The attempt at a solution The given choice of answers are in C, so I figured I had to convert the thermal expansion coefficients to C-1 but wasn't quite sure how to do that. K = C + 273 I figured the expansion would be Lo + [tex]\Delta[/tex]L I set that eq. for the Al and Cu bars equal, subbing in the [tex]\Delta[/tex]L with the equation above. My answer is off though, and i'm not sure if it's a matter of because I didn't convert my per Kelvin to per Celsius or what. I think it's an algebra thing but maybe i'm way off. QUESTION TWO: 1. The problem statement, all variables and given/known data The sound level at a point X is 14db below the sound level at a point 1.0m from a point source. The distance from the source to point X is: a) 2.0m b) 25m c) 5.0m d) 20.2m e) 4.0m 2. Relevant equations Decibels = 10 log (I/Io) Io = threshold of hearing = 1x10-12 W/m2 3. The attempt at a solution I don't know where to start with this one! QUESTION THREE: 1. The problem statement, all variables and given/known data A heat conducting rod, 1.60m long, is made of an Al section, 0.90m long, and a Cu section, 0.70m long. Both sections have a cross sectional area of 0.0004m2. The Al end and the copper end are maintained at temperatures of 30C and 170C respectively. The thermal conductivieis of Al and Cu are: kal = 205 W/m K and kcu = 385 W/m K. The rate at which heat is conducted is: a) 10W b) 9W c) 7.9W d) 12W E) 11W 2. Relevant equations Q = (kA[tex]\Delta[/tex]T)t/L 3. The attempt at a solution I couldn't figure out how to fit the t (time) into this at all, so perhaps i'm using the wrong equation? k values were all given A = 0.0004m2 [tex]\Delta[/tex]T = 170-30 = 140C Otherwise, I am also confused with this one. Do I need to find a seperate Q for both materials (Al and Cu)? And if so, what happens to that t (time) value? QUESTION FOUR: 1. The problem statement, all variables and given/known data A proton is fixed in place. A second proton is released from rest at a point 1.0cm away What will be the velocity of the second proton when it is a very large distance from the first? 2. Relevant equations Not sure, Wnc = 1/2 mvf2 ??? 3. The attempt at a solution Proton at rest = vo = 0m/s Proton mass = 1.67 x 10-27 kg QUESTION FIVE: 1. The problem statement, all variables and given/known data A bus is moving at 37.00m/s towards a wall. The sound from the bus has an original wavelength of 0.1500m. The sound from the bus reflects off the wall. What frequency sound does an observer on the moving bus hear from the reflection?? 2. Relevant equations Moving Observer: fo = fs (1 + vo/v) v = [tex]\lambda[/tex]f 3. The attempt at a solution Is this doppler effect?? I'm assuming the wavelength will double given that it reflects off a wall? So, [tex]\lambda[/tex] = 2(0.1500m) = 0.300m vo = 37.00m/s Since v = [tex]\lambda[/tex]f vi = [tex]\lambda[/tex]fi f = v / [tex]\lambda[/tex] = (343m/s)(0.1500) = 2286.667hz Subbing into equation: fo = (2286.667hz)(1 + 37m/s / 343m/s) = 2533hz Is that correct??? I don't think it is because doing that didn't account for the reflection off the wall - unless I misunderstand how that works? --------- Thanks for any help!! |
| Apr16-10, 07:31 PM | #2 |
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Bump.
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| Apr16-10, 09:17 PM | #3 |
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You have the right idea for question one by adding the original length to the expansion and then setting the two equations equal to each other. Find the temperature in kelvin first and then convert to Celsius, don't attempt to change the coefficients of thermal expansion.
A point source emits sound spherically. Intensity would therefore depend on the power dissipated over the surface area of the spherical wavefront. This one is pretty easy. They are asking for a rate so you divide by time. Also, by conservation of energy, the rates have to be the same. In the aluminum bar the temp. changes from 30 to some temp. T. The copper bar changes from T to 170. Convert to Kelvin. Set the equations equal. Solve for T. Solve for the rate. Simple conservation of energy problem. The proton experiences a repulsive force and there exists a potential energy at its initial position. "At infinity" there is no potential energy so all the energy has become kinetic. Solve for the speed. Wavelengths don't double when they reflect off walls. I don't know where you learned that. This is a Doppler shift problem. First the bus is a moving source emitting sound at some frequency (easy to solve for this). Find the frequency "heard" (incident) on the wall. The wall is now a source of sound waves of this new frequency. The bus is now a moving observer. Solve for the second Doppler shifted frequency. |
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