Originally posted by RamnedI'm going to make a bit of an assumption, explained below, that the wavelength coming out of the instrument is pretty much fixed. So now we're dealing with frequency = speed of sound / wavelength. Speed of sound increases as the square root of the absolute temperature. It's not a function of other air properties such as density and pressure, except as they change the temperature. So as the temperature increases, the speed of sound increases, and the frequency increases for a given wavelength. So as it gets hotter, you have to tune your instrument "lower" to get the same sound (pitch). Should be the same for string instruments, right?
Sound
[b]14. How does air temperature affect the tuning of a wind instrument?[/b]
(The wavelength of a note produced by the instrument in a given condition [positioning of the the valves] is essentially fixed--it is a function of the design of the instrument. Tuning changes this. So, eliminating the tuning, the wavelength is fixed. The temperature changes would have an extremely minimal and indirect effect. For example, the most extreme "normal" temperature changes, say, from 30 to 40 degrees F. to about 90 degrees F. would not expand or compress the instrument enough to change the wavelength produced. I'd have to know expansion coefficients and metal characteristics to be sure, so I'm making a bit of an assumption here. Plus, I'm guessing that mechanical design, hardening, alloying, and use of multiple metals all might contribute to reducing or eliminating this effect.)
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First off, generally, (in an uncompressed state) air pressure decreases as temperature increases. If the pressure is low, then sound waves travel faster. Since the speed of sound is increased, the frequency must be increased in proportion to the increase of the speed in sound, causing a higher pitch.
Feel free to correct me if I'm wrong.
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Originally posted by twilight2007You are correct. A rise in temperature will increase the dimensions of the wind instrument much less than it increases the speed of sound in the enclosed air. This effect will raise the resonant frequencies => sharp as temp increases, flat as it decreases.
First off, generally, (in an uncompressed state) air pressure decreases as temperature increases. If the pressure is low, then sound waves travel faster. Since the speed of sound is increased, the frequency must be increased in proportion to the increase of the speed in sound, causing a [b]higher pitch.
Feel free to correct me if I'm wrong.[/b]
Next question to follow. Good job. HolyT you have it the same - I misread "tune it lower."
Originally posted by twilight2007Isn't that what I said, except without the pressure business? Speed of sound (in a given medium) is a function of temperature, period. You can change the pressure all day and keep the temperature fixed, and the speed of sound stays the same. And I did say that the frequency (pitch) increases with increasing temp. Oh well. Keep 'em coming.
First off, generally, (in an uncompressed state) air pressure decreases as temperature increases. If the pressure is low, then sound waves travel faster. Since the speed of sound is increased, the frequency must be increased in proportion to the increase of the speed in sound, causing a [b]higher pitch.
Feel free to correct me if I'm wrong.[/b]
Originally posted by Ramned...
Electric Forces
[b]15. A balloon negatively charged by rubbing clings to a wall. (A) Is the wall positive charged or does it have a negative charge? (B) Why does the balloon eventually fall?[/b]
Negatives attracts positives? Therefore the wall is positively charged?
The negative charge only comes about by rubbing? Therefore...The negative charges would outnumber the positives for the time being?
Then the negative charges and the positive charges fall back into equilibrium, and the balloon falls away...
Correct?
Or is the wall neutrally charged and the balloon pulls out the positive charges by pushing back the negative charges?...
Am I right?
Originally posted by RamnedThe wall is negatively charged. More exactly the wall must have allmost zero charge but its outter surface certainly is negatively charged and that's what counts to us.
Electric Forces
[b]15. A balloon negatively charged by rubbing clings to a wall. (A) Is the wall positive charged or does it have a negative charge? (B) Why does the balloon eventually fall?[/b]
When we put the positevely charged balloon in contact with the negatively charged surface wall a potential difference will be created among them. That being the case the electrons of the wall will start to move from the wall to the balloon untill they are equally charged. At that point the potential difference ceases to exist and the ballon falls from the wall.
BTW The spring problem was a killer and after seeing both solutions I realised that I hadn't understood the problem correctly.
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Originally posted by adam warlockI nearly said the same thing. I never mentioned the potintial difference though. If that's the only thing I missed, (when I'm just beginning my AP Physics class), that's not too bad.
The wall is negatively charged. More exactly the wall must have allmost zero charge but its outter surface certainly is negatively charged and that's what counts to us.
When we put the positevely charged balloon in contact with the negatively charged surface wall a potential difference will be created among them. That being the case the electrons of th ...[text shortened]... ler and after seeing both solutions I realised that I hadn't understood the problem correctly.
The wall's outer surface is therefore positively charged? Or does the balloon pull forth protons?
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Originally posted by RamnedTwo o clock in here! 😳
The balloon is negatively charged when it clings to the wall.
Change the charges on the wall and balloon. The balloon is negative, the wall positive and the electrons flow from the ballon to the wall.
Sorry 😳
Edit: And the attraction comes about because of the electrons on the balloon repel the electrons on the wall. Sheesh!, no more late time answering in this thread for me.
Originally posted by EinsteinMindno 😉
I nearly said the same thing. I never mentioned the potintial difference though. If that's the only thing I missed, (when I'm just beginning my AP Physics class), that's not too bad.
The wall's outer surface is therefore positively charged? Or does the balloon pull forth protons?